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) $\frac { 3 } { 2 }$

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 $\_\_\_\_$.

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) $5 \sqrt { 3 }$

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

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

The lines $\frac { x - 2 } { 2 } = \frac { y } { - 2 } = \frac { z - 7 } { 16 }$ and $\frac { x + 3 } { 4 } = \frac { y + 2 } { 3 } = \frac { z + 2 } { 1 }$ intersect at the point $P$. If the distance of $P$ from the line $\frac { \mathrm { x } + 1 } { 2 } = \frac { \mathrm { y } - 1 } { 3 } = \frac { \mathrm { z } - 1 } { 1 }$ is $l$, then $14 l ^ { 2 }$ is equal to $\_\_\_\_$ .

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) $\frac { 6 \sqrt { 14 } } { 7 }$

Let the point $( - 1 , \alpha , \beta )$ lie on the line of the shortest distance between the lines $\frac { x + 2 } { - 3 } = \frac { y - 2 } { 4 } = \frac { z - 5 } { 2 }$ and $\frac { x + 2 } { - 1 } = \frac { y + 6 } { 2 } = \frac { z - 1 } { 0 }$. Then $( \alpha - \beta ) ^ { 2 }$ is equal to $\_\_\_\_$

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

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

Let $\mathrm { L } _ { 1 } : \frac { x - 1 } { 2 } = \frac { y - 2 } { 3 } = \frac { z - 3 } { 4 }$ and $\mathrm { L } _ { 2 } : \frac { x - 2 } { 3 } = \frac { y - 4 } { 4 } = \frac { z - 5 } { 5 }$ be two lines. Then which of the following points lies on the line of the shortest distance between $L _ { 1 }$ and $L _ { 2 }$?
(1) $\left( \frac { 14 } { 3 } , - 3 , \frac { 22 } { 3 } \right)$
(2) $\left( - \frac { 5 } { 3 } , - 7,1 \right)$
(3) $\left( 2,3 , \frac { 1 } { 3 } \right)$
(4) $\left( \frac { 8 } { 3 } , - 1 , \frac { 1 } { 3 } \right)$

The distance of the line $\frac { x - 2 } { 2 } = \frac { y - 6 } { 3 } = \frac { z - 3 } { 4 }$ from the point $( 1,4,0 )$ along the line $\frac { x } { 1 } = \frac { y - 2 } { 2 } = \frac { z + 3 } { 3 }$ is :
(1) $\sqrt { 17 }$
(2) $\sqrt { 15 }$
(3) $\sqrt { 14 }$
(4) $\sqrt { 13 }$

Q90. If the shortest distance between the lines $\frac { x + 2 } { 2 } = \frac { y + 3 } { 3 } = \frac { z - 5 } { 4 }$ and $\frac { x - 3 } { 1 } = \frac { y - 2 } { - 3 } = \frac { z + 4 } { 2 }$ is $\frac { 38 } { 3 \sqrt { 5 } } \mathrm { k }$, and $\int _ { 0 } ^ { \mathrm { k } } \left[ x ^ { 2 } \right] \mathrm { d } x = \alpha - \sqrt { \alpha }$, where $[ x ]$ denotes the greatest integer function, then $6 \alpha ^ { 3 }$ is equal to $\_\_\_\_$墐

ANSWER KEYS

\begin{tabular}{|l|l|} \hline 1. (3) & 2. (1) \hline 9. (3) & 10. (3) \hline 17. (1) & 18. (2) \hline 25. (16) & 26. (1) \hline 33. (4) & 34. (3) \hline 41. (4) & 42. (1) \hline 49. (4) & 50. (1) \hline 57. (100) & 58. (4) \hline 65. (1) & 66. (3) \hline

Q90. Let the point $( - 1 , \alpha , \beta )$ lie on the line of the shortest distance between the lines $\frac { x + 2 } { - 3 } = \frac { y - 2 } { 4 } = \frac { z - 5 } { 2 }$ and $\frac { x + 2 } { - 1 } = \frac { y + 6 } { 2 } = \frac { z - 1 } { 0 }$. Then $( \alpha - \beta ) ^ { 2 }$ is equal to $\_\_\_\_$

ANSWER KEYS

\begin{tabular}{|l|l|l|l|} \hline 1. (3) & 2. (3) & 3. (4) & 4. (2) \hline 9. (3) & 10. (4) & 11. (3) & 12. (2) \hline 17. (2) & 18. (2) & 19. (3) & 20. (2) \hline 25. (16) & 26. (5) & 27. (500) & 28. (4) \hline

Q89. If the shortest distance between the lines $\frac { x - \lambda } { 3 } = \frac { y - 2 } { - 1 } = \frac { z - 1 } { 1 }$ and $\frac { x + 2 } { - 3 } = \frac { y + 5 } { 2 } = \frac { z - 4 } { 4 }$ is $\frac { 44 } { \sqrt { 30 } }$, then the largest possible value of $| \lambda |$ is equal to $\_\_\_\_$

Q78. 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

Q79. 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

Q80. The shortest distance between the lines $\frac { x - 3 } { 4 } = \frac { y + 7 } { - 11 } = \frac { z - 1 } { 5 }$ and $\frac { x - 5 } { 3 } = \frac { y - 9 } { - 6 } = \frac { z + 2 } { 1 }$ is:
(1) $\frac { 178 } { \sqrt { 563 } }$
(2) $\frac { 187 } { \sqrt { 563 } }$
(3) $\frac { 185 } { \sqrt { 563 } }$
(4) $\frac { 179 } { \sqrt { 563 } }$

In tetrahedron $A B C D$, $\overline { A B } = \overline { A C } = \overline { A D } = 4 \sqrt { 6 }$, $\overline { B D } = \overline { C D } = 8$, and $\cos \angle B A C = \frac { 1 } { 3 }$. The distance from point $D$ to plane $A B C$ is (31) $\sqrt { (32) }$. (Express as a fraction in simplest radical form)

In coordinate space, there are two non-intersecting lines $L_{1}: \left\{\begin{array}{l} x = 1+t \\ y = 1-t \\ z = 2+t \end{array}\right.$, $t$ is a real number, $L_{2}: \left\{\begin{array}{l} x = 2+2s \\ y = 5+s \\ z = 6-s \end{array}\right.$, $s$ is a real number. Another line $L_{3}$ intersects both $L_{1}$ and $L_{2}$ and is perpendicular to both. If points $P$ and $Q$ are on $L_{1}$ and $L_{2}$ respectively and both are at distance 3 from $L_{3}$, then the distance between points $P$ and $Q$ is (17--1)$\sqrt{17\text{-}2}$. (Express as a simplified radical)

In coordinate space, point $A$ has coordinates $(a, b, c)$, where $a, b, c$ are all negative real numbers. Point $A$ is at distance 6 from each of the three planes $E _ { 1 } : 4 y + 3 z = 2$, $E _ { 2 } : 3 y + 4 z = - 5$, and $E _ { 3 } : x + 2 y + 2 z = - 2$. Then $a + b + c =$ (14-1) (14-2) (14-3).