Consider the lines L₁ : (x+1)/3 =...

MathematicsShortest Distance Between Two Skew LinesJEE Advanced 2008Moderate

Comprehension Passage

Consider the lines

L_{1} : \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2}

and

L_{2} : \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}

Question 1:

The unit vector perpendicular to both $L_{1}$ and $L_{2}$ is

(A)

\frac{- i ^ + 7 j ^ + 7 k ^}{99}

(B)

\frac{- i ^ - 7 j ^ + 5 k ^}{5 3}

(C)

\frac{- i ^ + 7 j ^ + 5 k ^}{5 3}

(D)

\frac{7 i ^ - 7 j ^ - k ^}{99}

Answer:B

Question 2:

The shortest distance between $L_{1}$ and $L_{2}$ is

(A)

(B)

\frac{17}{3}

(C)

\frac{41}{5 3}

(D)

\frac{17}{5 3}

Answer:D

Question 3:

The distance of the point (1, 1, 1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines $L_{1}$ and $L_{2}$ is

(A)

\frac{2}{75}

(B)

\frac{7}{75}

(C)

\frac{13}{75}

(D)

\frac{23}{75}

Answer:C

Visualized Solution (English)

Identify Direction Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b_{1} and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b_{2}

Given lines: $L_{1} : \frac{x + 1}{3} = \frac{y + 2}{1} = \frac{z + 1}{2}$ and $L_{2} : \frac{x - 2}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}$
Direction vector of $L_{1}$ : $b_{1} = 3 \hat{i} + \hat{j} + 2 \hat{k}$
Direction vector of $L_{2}$ : $b_{2} = \hat{i} + 2 \hat{j} + 3 \hat{k}$

Calculate Cross Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b_{1} \times b_{2}

To find a vector $n$ perpendicular to both $L_{1}$ and $L_{2}$ , we use the cross product:
$n = b_{1} \times b_{2} = \hat{i} 31 \hat{j} 12 \hat{k} 23$
$n = \hat{i} (3 - 4) - \hat{j} (9 - 2) + \hat{k} (6 - 1) = - \hat{i} - 7 \hat{j} + 5 \hat{k}$

Find the Unit Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \overset{n}{^}

Magnitude of $n$ : $∣ n ∣ = (- 1)^{2} + (- 7)^{2} + 5^{2} = 1 + 49 + 25 = 75 = 53$
The unit vector $\overset{n}{^} = \frac{n}{∣ n ∣} = \frac{- i ^ - 7 j ^ + 5 k ^}{5 3}$
This matches option (b) for the first sub-question.

Identify Points .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A_{1} and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A_{2} on the Lines

Point on $L_{1}$ : $A_{1} (- 1, - 2, - 1) \Rightarrow a_{1} = - \hat{i} - 2 \hat{j} - \hat{k}$
Point on $L_{2}$ : $A_{2} (2, - 2, 3) \Rightarrow a_{2} = 2 \hat{i} - 2 \hat{j} + 3 \hat{k}$

Calculate Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a_{2} - a_{1}

Vector connecting points: $a_{2} - a_{1} = (2 - (- 1)) \hat{i} + (- 2 - (- 2)) \hat{j} + (3 - (- 1)) \hat{k}$
$a_{2} - a_{1} = 3 \hat{i} + 0 \hat{j} + 4 \hat{k}$

Shortest Distance Calculation

Shortest distance $d = \frac{∣ ( a _{2} - a _{1} ) \cdot ( b _{1} \times b _{2} ) ∣}{∣ b _{1} \times b _{2} ∣}$
Numerator: $∣ (3 \hat{i} + 4 \hat{k}) \cdot (- \hat{i} - 7 \hat{j} + 5 \hat{k}) ∣ = ∣ - 3 + 0 + 20∣ = 17$
Denominator: $53$
$d = \frac{17}{5 3}$ . This matches option (d) for the second sub-question.

Equation of the Plane

Plane passes through $A_{1} (- 1, - 2, - 1)$ with normal $n = - \hat{i} - 7 \hat{j} + 5 \hat{k}$
Equation: $- 1 (x + 1) - 7 (y + 2) + 5 (z + 1) = 0$
$- x - 1 - 7 y - 14 + 5 z + 5 = 0 \Rightarrow - x - 7 y + 5 z - 10 = 0$
Simplified Equation: $x + 7 y - 5 z + 10 = 0$

Distance of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1, 1, 1) from Plane

Point $P (1, 1, 1)$ , Plane $x + 7 y - 5 z + 10 = 0$
Distance $D = \frac{∣1 ( 1 ) + 7 ( 1 ) - 5 ( 1 ) + 10∣}{1 ^{2} + 7 ^{2} + ( - 5 ) ^{2}}$
$D = \frac{∣1 + 7 - 5 + 10∣}{1 + 49 + 25} = \frac{13}{75}$
This matches option (c) for the third sub-question.

Summary and Key Takeaways

Key Takeaways:
The cross product $b_{1} \times b_{2}$ provides the direction of the common perpendicular.
Shortest distance is the projection of any vector connecting the lines onto the common perpendicular.
The equation of a plane requires a point and a normal vector.
Next Challenge: Prove that if the shortest distance is zero, the lines are coplanar.

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Conceptually Similar Problems

MathematicsEquation of a Line in SpaceJEE Advanced 2013Moderate

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A line $l$ passing through the origin is perpendicular to the lines $l_{1} : (3 + t) \hat{i} + (- 1 + 2 t) \hat{j} + (4 + 2 t) \hat{k}, - \infty < t < \infty$ and $l_{2} : (3 + 2 s) \hat{i} + (3 + 2 s) \hat{j} + (2 + s) \hat{k}, - \infty < s < \infty$ . Then, the coordinate(s) of the point(s) on $l_{2}$ at a distance of $17$ from the point of intersection of $l$ and $l_{1}$ is (are)

* Multiple Correct Options

(A)

(\frac{7}{3}, \frac{7}{3}, \frac{5}{3})

(B)

(- 1, - 1, 0)

(C)

(1, 1, 1)

(D)

(\frac{7}{9}, \frac{7}{9}, \frac{8}{9})

Answer:B, D

MathematicsEquation of a PlaneJEE Advanced 2006Moderate

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A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

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0

(B)

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In $R^{3}$ , let $L$ be a straight line passing through the origin. Suppose that all the points on $L$ are at a constant distance from the two planes $P_{1} : x + 2 y - z + 1 = 0$ and $P_{2} : 2 x - y + z - 1 = 0$ . Let $M$ be the locus of the feet of the perpendiculars drawn from the points on $L$ to the plane $P_{1}$ . Which of the following points lie(s) on $M$ ?

* Multiple Correct Options

(A)

(0, - \frac{5}{6}, - \frac{2}{3})

(B)

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(C)

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(D)

(- \frac{1}{3}, 0, \frac{2}{3})

Answer:A, B

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Consider the lines $L_{1} : \frac{x - 1}{2} = \frac{y}{- 1} = \frac{z + 3}{1}, L_{2} : \frac{x - 4}{1} = \frac{y + 3}{1} = \frac{z + 3}{2}$ and the planes $P_{1} : 7 x + y + 2 z = 3, P_{2} : 3 x + 5 y - 6 z = 4$ . Let $a x + b y + cz = d$ be the equation of the plane passing through the point of intersection of lines $L_{1}$ and $L_{2}$ , and perpendicular to planes $P_{1}$ and $P_{2}$ . Match List I with List II:

List-I

(P)

(Q)

(R)

(S)

List-II

(1)

(2)

-3

(3)

(4)

-2

Answer:P → 3, Q → 2, R → 4, S → 1

MathematicsEquation of a PlaneJEE Advanced 2010Moderate

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If the distance between the plane $A x - 2 y + z = d$ and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is $6$ , then find $∣ d ∣$ .

Answer:6

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In $R^{3}$ , consider the planes $P_{1} : y = 0$ and $P_{2} : x + z = 1$ . Let $P_{3}$ be the plane, different from $P_{1}$ and $P_{2}$ , which passes through the intersection of $P_{1}$ and $P_{2}$ . If the distance of the point $(0, 1, 0)$ from $P_{3}$ is 1 and the distance of a point $(α, β, γ)$ from $P_{3}$ is 2, then which of the following relations is (are) true?

* Multiple Correct Options

(A)

2 α + β + 2 γ + 2 = 0

(B)

2 α - β + 2 γ + 4 = 0

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MathematicsIntersection of a Line and a PlaneJEE Main 2015Easy

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The distance of the point $(1, 0, 2)$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ , is

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321

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The equation of a plane passing through the line of intersection of the planes $x + 2 y + 3 z = 2$ and $x - y + z = 3$ and at a distance $\frac{2}{3}$ from the point $(3, 1, - 1)$ is

(A)

5 x - 11 y + z = 17

(B)

2 x + y = 32 - 1

(C)

x + y + z = 3

(D)

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Answer:A

MathematicsIntersection of a Line and a PlaneJEE Advanced 2010Easy

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If the distance of the point $P (1, - 2, 1)$ from the plane $x + 2 y - 2 z = α$ , where $α > 0$ , is 5, then the foot of the perpendicular from $P$ to the plane is

(A)

(\frac{8}{3}, \frac{4}{3}, - \frac{7}{3})

(B)

(\frac{4}{3}, \frac{4}{3}, \frac{1}{3})

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(\frac{1}{3}, \frac{2}{3}, \frac{10}{3})

(D)

(\frac{2}{3}, \frac{1}{3}, \frac{5}{2})

Answer:A

MathematicsEquation of a PlaneJEE Advanced 2004Moderate

View in:English Hindi

Find the equation of plane passing through $(1, 1, 1)$ & parallel to the lines $L_{1}, L_{2}$ having direction ratios $(1, 0, - 1), (1, - 1, 0)$ . Find the volume of tetrahedron formed by origin and the points where these planes intersect the coordinate axes.

Answer:

x + y + z = 3

;

\frac{9}{2}

cubic units

Comprehension Passage

The unit vector perpendicular to both $L_{1}$ and $L_{2}$ is

The shortest distance between $L_{1}$ and $L_{2}$ is

The distance of the point (1, 1, 1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines $L_{1}$ and $L_{2}$ is

Visualized Solution (English)

Identify Direction Vectors $b_{1}$ and $b_{2}$

Calculate Cross Product $b_{1} \times b_{2}$

Find the Unit Vector $\overset{n}{^}$

Identify Points $A_{1}$ and $A_{2}$ on the Lines

Calculate Vector $a_{2} - a_{1}$

Shortest Distance Calculation

Equation of the Plane

Distance of Point $(1, 1, 1)$ from Plane

Summary and Key Takeaways

Conceptually Similar Problems

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

List-I

List-II

If the distance between the plane $A x - 2 y + z = d$ and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is $6$ , then find $∣ d ∣$ .

The distance of the point $(1, 0, 2)$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ , is

The equation of a plane passing through the line of intersection of the planes $x + 2 y + 3 z = 2$ and $x - y + z = 3$ and at a distance $\frac{2}{3}$ from the point $(3, 1, - 1)$ is

If the distance of the point $P (1, - 2, 1)$ from the plane $x + 2 y - 2 z = α$ , where $α > 0$ , is 5, then the foot of the perpendicular from $P$ to the plane is

Find the equation of plane passing through $(1, 1, 1)$ & parallel to the lines $L_{1}, L_{2}$ having direction ratios $(1, 0, - 1), (1, - 1, 0)$ . Find the volume of tetrahedron formed by origin and the points where these planes intersect the coordinate axes.

Comprehension Passage

The unit vector perpendicular to both $L_{1}$ and $L_{2}$ is

The shortest distance between $L_{1}$ and $L_{2}$ is

The distance of the point (1, 1, 1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines $L_{1}$ and $L_{2}$ is

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

List-I

List-II

If the distance between the plane $A x - 2 y + z = d$ and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is $6$ , then find $∣ d ∣$ .

The distance of the point $(1, 0, 2)$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ , is

The equation of a plane passing through the line of intersection of the planes $x + 2 y + 3 z = 2$ and $x - y + z = 3$ and at a distance $\frac{2}{3}$ from the point $(3, 1, - 1)$ is

If the distance of the point $P (1, - 2, 1)$ from the plane $x + 2 y - 2 z = α$ , where $α > 0$ , is 5, then the foot of the perpendicular from $P$ to the plane is

Find the equation of plane passing through $(1, 1, 1)$ & parallel to the lines $L_{1}, L_{2}$ having direction ratios $(1, 0, - 1), (1, - 1, 0)$ . Find the volume of tetrahedron formed by origin and the points where these planes intersect the coordinate axes.

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The unit vector perpendicular to both L1​ and L2​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The shortest distance between L1​ and L2​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the point (1, 1, 1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines L1​ and L2​ is

Visualized Solution (English)

Identify Direction Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b1​​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2​​

Calculate Cross Product .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b1​​×b2​​

Find the Unit Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n^

Identify Points .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A2​ on the Lines

Calculate Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a2​​−a1​​

Shortest Distance Calculation

Equation of the Plane

Distance of Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (1,1,1) from Plane

Summary and Key Takeaways

Conceptually Similar Problems

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A plane which is perpendicular to two planes 2x−2y+z=0 and x−y+2z=4, passes through (1,−2,1). The distance of the plane from the point (1,2,2) is

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the distance between the plane Ax−2y+z=d and the plane containing the lines 2x−1​=3y−2​=4z−3​ and 3x−2​=4y−3​=5z−4​ is 6​, then find ∣d∣.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the point (1,0,2) from the point of intersection of the line 3x−2​=4y+1​=12z−2​ and the plane x−y+z=16, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of a plane passing through the line of intersection of the planes x+2y+3z=2 and x−y+z=3 and at a distance 3​2​ from the point (3,1,−1) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the distance of the point P(1,−2,1) from the plane x+2y−2z=α, where α>0, is 5, then the foot of the perpendicular from P to the plane is

Comprehension Passage

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The unit vector perpendicular to both L1​ and L2​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The shortest distance between L1​ and L2​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the point (1, 1, 1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines L1​ and L2​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A plane which is perpendicular to two planes 2x−2y+z=0 and x−y+2z=4, passes through (1,−2,1). The distance of the plane from the point (1,2,2) is

List-I

List-II

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the distance between the plane Ax−2y+z=d and the plane containing the lines 2x−1​=3y−2​=4z−3​ and 3x−2​=4y−3​=5z−4​ is 6​, then find ∣d∣.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the point (1,0,2) from the point of intersection of the line 3x−2​=4y+1​=12z−2​ and the plane x−y+z=16, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The equation of a plane passing through the line of intersection of the planes x+2y+3z=2 and x−y+z=3 and at a distance 3​2​ from the point (3,1,−1) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the distance of the point P(1,−2,1) from the plane x+2y−2z=α, where α>0, is 5, then the foot of the perpendicular from P to the plane is

The unit vector perpendicular to both $L_{1}$ and $L_{2}$ is

The shortest distance between $L_{1}$ and $L_{2}$ is

The distance of the point (1, 1, 1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines $L_{1}$ and $L_{2}$ is

Identify Direction Vectors $b_{1}$ and $b_{2}$

Calculate Cross Product $b_{1} \times b_{2}$

Find the Unit Vector $\overset{n}{^}$

Identify Points $A_{1}$ and $A_{2}$ on the Lines

Calculate Vector $a_{2} - a_{1}$

Distance of Point $(1, 1, 1)$ from Plane

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

If the distance between the plane $A x - 2 y + z = d$ and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is $6$ , then find $∣ d ∣$ .

The distance of the point $(1, 0, 2)$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ , is

The equation of a plane passing through the line of intersection of the planes $x + 2 y + 3 z = 2$ and $x - y + z = 3$ and at a distance $\frac{2}{3}$ from the point $(3, 1, - 1)$ is

If the distance of the point $P (1, - 2, 1)$ from the plane $x + 2 y - 2 z = α$ , where $α > 0$ , is 5, then the foot of the perpendicular from $P$ to the plane is

The unit vector perpendicular to both $L_{1}$ and $L_{2}$ is

The shortest distance between $L_{1}$ and $L_{2}$ is

The distance of the point (1, 1, 1) from the plane passing through the point (-1, -2, -1) and whose normal is perpendicular to both the lines $L_{1}$ and $L_{2}$ is

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

If the distance between the plane $A x - 2 y + z = d$ and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is $6$ , then find $∣ d ∣$ .

The distance of the point $(1, 0, 2)$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 16$ , is

The equation of a plane passing through the line of intersection of the planes $x + 2 y + 3 z = 2$ and $x - y + z = 3$ and at a distance $\frac{2}{3}$ from the point $(3, 1, - 1)$ is

If the distance of the point $P (1, - 2, 1)$ from the plane $x + 2 y - 2 z = α$ , where $α > 0$ , is 5, then the foot of the perpendicular from $P$ to the plane is