Let $N$ be the foot of perpendicular from the point $P ( 1 , - 2 , 3 )$ on the line passing through the points $( 4 , 5 , 8 )$ and $( 1 , - 7 , 5 )$. Then the distance of $N$ from the plane $2x - 2y + z + 5 = 0$ is $\_\_\_\_$.

Let $S$ be the set of all values of $\lambda$, for which the shortest distance between the lines $\frac { x - \lambda } { 0 } = \frac { y - 3 } { 4 } = \frac { z + 6 } { 1 }$ and $\frac { x + \lambda } { 3 } = \frac { y } { - 4 } = \frac { z - 6 } { 0 }$ is 13. Then $8 \left| \sum _ { \lambda \in S } \lambda \right|$ is equal to
(1) 306
(2) 304
(3) 308
(4) 302

The shortest distance between the lines $x + 1 = 2 y = - 12 z$ and $x = y + 2 = 6 z - 6$ is
(1) 2
(2) 3
(3) $\frac { 5 } { 2 }$
(4) $\frac { 3 } { 2 }$

Let the plane $x + 3 y - 2 z + 6 = 0$ meet the co-ordinate axes at the points $A , B , C$. If the orthocenter of the triangle $A B C$ is $\left( \alpha , \beta , \frac { 6 } { 7 } \right)$, then $98 ( \alpha + \beta ) ^ { 2 }$ is equal to $\_\_\_\_$ .

The foot of perpendicular of the point $( 2,0,5 )$ on the line $\frac { x + 1 } { 2 } = \frac { y - 1 } { 5 } = \frac { z + 1 } { - 1 }$ is $( \alpha , \beta , \gamma )$. Then, which of the following is NOT correct?
(1) $\frac { \alpha \beta } { \gamma } = \frac { 4 } { 15 }$
(2) $\frac { \alpha } { \beta } = - 8$
(3) $\frac { \beta } { \gamma } = - 5$
(4) $\frac { \gamma } { \alpha } = \frac { 5 } { 8 }$

Let the equation of the plane P containing the line $x + 10 = \frac { 8 - y } { 2 } = z$ be $a x + b y + 3 z = 2 ( a + b )$ and the distance of the plane P from the point $( 1,27,7 )$ be $c$. Then $\mathrm { a } ^ { 2 } + \mathrm { b } ^ { 2 } + \mathrm { c } ^ { 2 }$ is equal to

Shortest distance between the lines $\frac { x - 1 } { 2 } = \frac { y + 8 } { - 7 } = \frac { z - 4 } { 5 }$ and $\frac { x - 1 } { 2 } = \frac { y - 2 } { 1 } = \frac { z - 6 } { - 3 }$ is (1) $2 \sqrt { 3 }$ (2) $4 \sqrt { 3 }$ (3) $3 \sqrt { 3 }$ (4) $5 \sqrt { 3 }$

Let the lines $L _ { 1 } : \frac { x + 5 } { 3 } = \frac { y + 4 } { 1 } = \frac { z - \alpha } { - 2 }$ and $L _ { 2 } : 3 x + 2 y + z - 2 = 0 = x - 3 y + 2 z - 13$ be coplanar. If the point $P ( a , b , c )$ on $L _ { 1 }$ is nearest to the point $Q ( - 4 , - 3,2 )$, then $| a | + | b | + | c |$ is equal to
(1) 12
(2) 14
(3) 8
(4) 10

If the shortest distance between the line joining the points $( 1,2,3 )$ and $( 2,3,4 )$, and the line $\frac { x - 1 } { 2 } = \frac { y + 1 } { - 1 } = \frac { z - 2 } { 0 }$ is $\alpha$, then $28 \alpha ^ { 2 }$ is equal to $\_\_\_\_$.

The plane $2 x - y + z = 4$ intersects the line segment joining the points $\mathrm { A } ( \mathrm { a } , - 2,4 )$ and $\mathrm { B } ( 2 , \mathrm {~b} , - 3 )$ at the point C in the ratio 2 : 1 and the distance of the point C from the origin is $\sqrt { 5 }$. If $a b < 0$ and P is the point $( \mathrm { a } - \mathrm { b } , \mathrm { b } , 2 \mathrm {~b} - \mathrm { a } )$ then $\mathrm { CP } ^ { 2 }$ is equal to: (1) $\frac { 17 } { 3 }$ (2) $\frac { 16 } { 3 }$ (3) $\frac { 73 } { 3 }$ (4) $\frac { 97 } { 3 }$

Let the plane $P : 4 x - y + z = 10$ be rotated by an angle $\frac { \pi } { 2 }$ about its line of intersection with the plane $x + y - z = 4$. If $\alpha$ is the distance of the point $( 2,3 , - 4 )$ from the new position of the plane $P$, then $35 \alpha$ is equal to
(1) 85
(2) 105
(3) 126
(4) 90

Let the equation of the plane passing through the line $x - 2 y - z - 5 = 0 = x + y + 3 z - 5$ and parallel to the line $x + y + 2 z - 7 = 0 = 2 x + 3 y + z - 2$ be $a x + b y + c z = 65$. Then the distance of the point $( a , b , c )$ from the plane $2 x + 2 y - z + 16 = 0$ is $\_\_\_\_$.

If the lines $\frac { x - 1 } { 1 } = \frac { y - 2 } { 2 } = \frac { z + 3 } { 1 }$ and $\frac { x - a } { 2 } = \frac { y + 2 } { 3 } = \frac { z - 3 } { 1 }$ intersects at the point $P$, then the distance of the point $P$ from the plane $z = a$ is : (1) 16 (2) 28 (3) 10 (4) 22

If the mirror image of the point $P(3,4,9)$ in the line $\frac{x-1}{3} = \frac{y+1}{2} = \frac{z-2}{1}$ is $(\alpha, \beta, \gamma)$, then $14\alpha + \beta + \gamma$ is:
(1) 102
(2) 138
(3) 108
(4) 132

If the shortest distance between the lines $\begin{aligned} & L _ { 1 } : \vec { r } = ( 2 + \lambda ) \hat { i } + ( 1 - 3 \lambda ) \hat { j } + ( 3 + 4 \lambda ) \hat { k } , \quad \lambda \in \mathbb { R } \\ & L _ { 2 } : \vec { r } = 2 ( 1 + \mu ) \hat { i } + 3 ( 1 + \mu ) \hat { j } + ( 5 + \mu ) \hat { k } , \quad \mu \in \mathbb { R } \end{aligned}$ is $\frac { m } { \sqrt { n } }$, where $\operatorname { gcd } ( m , n ) = 1$, then the value of $m + n$ equals
(1) 390
(2) 384
(3) 377
(4) 387

The shortest distance between the lines $\frac { x - 3 } { 2 } = \frac { y + 15 } { - 7 } = \frac { z - 9 } { 5 }$ and $\frac { x + 1 } { 2 } = \frac { y - 1 } { 1 } = \frac { z - 9 } { - 3 }$ is
(1) $8 \sqrt { 3 }$
(2) $4 \sqrt { 3 }$
(3) $5 \sqrt { 3 }$
(4) $6 \sqrt { 3 }$

The distance, of the point $( 7 , - 2,11 )$ from the line $\frac { x - 6 } { 1 } = \frac { y - 4 } { 0 } = \frac { z - 8 } { 3 }$ along the line $\frac { x - 5 } { 2 } = \frac { y - 1 } { - 3 } = \frac { z - 5 } { 6 }$, is:
(1) 12
(2) 14
(3) 18
(4) 21

Let $P Q R$ be a triangle with $R ( - 1,4,2 )$. Suppose $M ( 2,1,2 )$ is the mid point of $P Q$. The distance of the centroid of $\triangle P Q R$ from the point of intersection of the line $\frac { x - 2 } { 0 } = \frac { y } { 2 } = \frac { z + 3 } { - 1 }$ and $\frac { x - 1 } { 1 } = \frac { y + 3 } { - 3 } = \frac { z + 1 } { 1 }$ is
(1) 69
(2) 9
(3) $\sqrt { 69 }$
(4) $\sqrt { 99 }$

Let the image of the point $( 1,0,7 )$ in the line $\frac { x } { 1 } = \frac { y - 1 } { 2 } = \frac { z - 2 } { 3 }$ be the point $( \alpha , \beta , \gamma )$. Then which one of the following points lies on the line passing through $( \alpha , \beta , \gamma )$ and making angles $\frac { 2 \pi } { 3 }$ and $\frac { 3 \pi } { 4 }$ with y - axis and z axis respectively and an acute angle with x -axis?
(1) $( 1 , - 2,1 + \sqrt { 2 } )$
(2) $( 1,2,1 - \sqrt { 2 } )$
(3) $( 3,4,3 - 2 \sqrt { 2 } )$
(4) $( 3 , - 4,3 + 2 \sqrt { 2 } )$

Let $( \alpha , \beta , \gamma )$ be the image of the point $( 8,5,7 )$ in the line $\frac { x - 1 } { 2 } = \frac { y + 1 } { 3 } = \frac { z - 2 } { 5 }$. Then $\alpha + \beta + \gamma$ is equal to :
(1) 16
(2) 20
(3) 14
(4) 18

If the shortest distance between the lines $\frac { x - \lambda } { 2 } = \frac { y - 4 } { 3 } = \frac { z - 3 } { 4 }$ and $\frac { x - 2 } { 4 } = \frac { y - 4 } { 6 } = \frac { z - 7 } { 8 }$ is $\frac { 13 } { \sqrt { 29 } }$, then a value of $\lambda$ is : (1) - 1 (2) $- \frac { 13 } { 25 }$ (3) $\frac { 13 } { 25 }$ (4) 1

Consider the line $L$ passing through the points $( 1,2,3 )$ and $( 2,3,5 )$. The distance of the point $\left( \frac { 11 } { 3 } , \frac { 11 } { 3 } , \frac { 19 } { 3 } \right)$ from the line L along the line $\frac { 3 x - 11 } { 2 } = \frac { 3 y - 11 } { 1 } = \frac { 3 z - 19 } { 2 }$ is equal to
(1) 6
(2) 5
(3) 4
(4) 3

If the shortest distance between the lines $\frac { x - 4 } { 1 } = \frac { y + 1 } { 2 } = \frac { z } { - 3 }$ and $\frac { x - \lambda } { 2 } = \frac { y + 1 } { 4 } = \frac { z - 2 } { - 5 }$ is $\frac { 6 } { \sqrt { 5 } }$, then the sum of all possible values of $\lambda$ is:
(1) 5
(2) 8
(3) 7
(4) 10

Let P be the point of intersection of the lines $\frac { x - 2 } { 1 } = \frac { y - 4 } { 5 } = \frac { z - 2 } { 1 }$ and $\frac { x - 3 } { 2 } = \frac { y - 2 } { 3 } = \frac { z - 3 } { 2 }$. Then, the shortest distance of P from the line $4 x = 2 y = z$ is
(1) $\frac { 5 \sqrt { 14 } } { 7 }$
(2) $\frac { 3 \sqrt { 14 } } { 7 }$
(3) $\frac { \sqrt { 14 } } { 7 }$
(4) $\frac { 6 \sqrt { 14 } } { 7 }$

Consider a line L passing through the points $\mathrm { P } ( 1,2,1 )$ and $\mathrm { Q } ( 2,1 , - 1 )$. If the mirror image of the point $\mathrm { A } ( 2,2,2 )$ in the line L is $( \alpha , \beta , \gamma )$, then $\alpha + \beta + 6 \gamma$ is equal to $\_\_\_\_$