A sphere with center $\mathrm { C } ( 0,1,1 )$ and radius $2 \sqrt { 2 }$ intersects the line $\frac { x } { 2 } = y = - z$ at two points A and B. Let $S$ be the area of triangle CAB. Find the value of $S ^ { 2 }$.

On plane $\alpha$, there is a right isosceles triangle ABC with $\angle \mathrm { A } = 90 ^ { \circ }$ and $\overline { \mathrm { BC } } = 6$. A point P outside plane $\alpha$ is at a distance of 4 from the plane, and the foot of the perpendicular from P to plane $\alpha$ is point A. What is the distance from point P to line BC? [3 points]
(1) $3 \sqrt { 2 }$
(2) 5
(3) $3 \sqrt { 3 }$
(4) $4 \sqrt { 2 }$
(5) 6

Let $l$ be the line passing through two distinct points $\mathrm { A } , \mathrm { B }$ on plane $\alpha$, and let H be the foot of the perpendicular from point P (not on plane $\alpha$) to plane $\alpha$. When $\overline { \mathrm { AB } } = \overline { \mathrm { PA } } = \overline { \mathrm { PB } } = 6 , \overline { \mathrm { PH } } = 4$, what is the distance between point H and line $l$? [3 points]
(1) $\sqrt { 11 }$
(2) $2 \sqrt { 3 }$
(3) $\sqrt { 13 }$
(4) $\sqrt { 14 }$
(5) $\sqrt { 15 }$

From a point $P(\lambda, \lambda, \lambda)$, perpendiculars $PQ$ and $PR$ are drawn respectively on the lines $y = x, z = 1$ and $y = -x, z = -1$. If $P$ is such that $\angle QPR$ is a right angle, then the possible value(s) of $\lambda$ is(are)
(A) $\sqrt{2}$
(B) $1$
(C) $-1$
(D) $-\sqrt{2}$

The distance of the point $(1, 3, -7)$ from the plane passing through the point $(1, -1, -1)$, having normal perpendicular to both the lines $\dfrac{x-1}{1} = \dfrac{y+2}{-2} = \dfrac{z-4}{3}$ and $\dfrac{x-2}{2} = \dfrac{y+1}{-1} = \dfrac{z+7}{-1}$, is:
(1) $\dfrac{20}{\sqrt{74}}$
(2) $\dfrac{10}{\sqrt{83}}$
(3) $\dfrac{5}{\sqrt{83}}$
(4) $\dfrac{10}{\sqrt{74}}$

The length of the projection of the line segment joining the points $( 5 , - 1,4 )$ and $( 4 , - 1,3 )$ on the plane, $x + y + z = 7$ is
(1) $\sqrt { \frac { 2 } { 3 } }$
(2) $\frac { 2 } { \sqrt { 3 } }$
(3) $\frac { 2 } { 3 }$
(4) $\frac { 1 } { 3 }$

The vertices $B$ and $C$ of a $\triangle A B C$ lie on the line, $\frac { x + 2 } { 3 } = \frac { y - 1 } { 0 } = \frac { z } { 4 }$ such that $B C = 5$ units. Then the area (in sq. units) of this triangle, given the point $A ( 1 , - 1,2 )$, is
(1) 6
(2) $2 \sqrt { 34 }$
(3) $\sqrt { 34 }$
(4) $5 \sqrt { 17 }$

If for $a > 0$, the feet of perpendiculars from the points $A ( a , - 2 a , 3 )$ and $B ( 0,4,5 )$ on the plane $l x + m y + n z = 0$ are points $C ( 0 , - a , - 1 )$ and $D$ respectively, then the length of line segment $C D$ is equal to :
(1) $\sqrt { 31 }$
(2) $\sqrt { 41 }$
(3) $\sqrt { 55 }$
(4) $\sqrt { 66 }$

Let $S$ be the mirror image of the point $Q ( 1,3,4 )$ with respect to the plane $2 x - y + z + 3 = 0$ and let $R ( 3,5 , \gamma )$ be a point of this plane. Then the square of the length of the line segment $S R$ is

Let $Q$ be the foot of perpendicular drawn from the point $P(1, 2, 3)$ to the plane $x + 2y + z = 14$. If $R$ is a point on the plane such that $\angle PRQ = 60^\circ$, then the area of $\triangle PQR$ is equal to

The length of the perpendicular from the point $( 1 , - 2,5 )$ on the line passing through $( 1,2,4 )$ and parallel to the line $x + y - z = 0 = x - 2 y + 3 z - 5$ is:
(1) $\sqrt { \frac { 21 } { 2 } }$
(2) $\sqrt { \frac { 9 } { 2 } }$
(3) $\sqrt { \frac { 73 } { 2 } }$
(4) 1

Let $Q$ and $R$ be two points on the line $\frac { x + 1 } { 2 } = \frac { y + 2 } { 3 } = \frac { z - 1 } { 2 }$ at a distance $\sqrt { 26 }$ from the point $P ( 4,2,7 )$. Then the square of the area of the triangle $PQR$ is $\_\_\_\_$.

If the length of the perpendicular drawn from the point $P ( a , 4,2 ) , a > 0$ on the line $\frac { x + 1 } { 2 } = \frac { y - 3 } { 3 } = \frac { z - 1 } { - 1 }$ is $2 \sqrt { 6 }$ units and $Q \left( \alpha _ { 1 } , \alpha _ { 2 } , \alpha _ { 3 } \right)$ is the image of the point $P$ in this line, then $a + \sum _ { i = 1 } ^ { 3 } \alpha _ { i }$ is equal to
(1) 7
(2) 8
(3) 12
(4) 14

The distance of the point $P ( 4,6 , - 2 )$ from the line passing through the point $( - 3,2,3 )$ and parallel to a line with direction ratios $3,3 , - 1$ is equal to:
(1) 3
(2) $\sqrt { 6 }$
(3) $2 \sqrt { 3 }$
(4) $\sqrt { 14 }$

Let the co-ordinates of one vertex of $\triangle A B C$ be $A ( 0,2 , \alpha )$ and the other two vertices lie on the line $\frac { \mathrm { x } + \alpha } { 5 } = \frac { \mathrm { y } - 1 } { 2 } = \frac { \mathrm { z } + 4 } { 3 }$. For $\alpha \in \mathbb { Z }$, if the area of $\triangle A B C$ is 21 sq. units and the line segment $B C$ has length $2 \sqrt { 21 }$ units, then $\alpha ^ { 2 }$ is equal to $\_\_\_\_$ .

Let a line $L$ pass through the point $P(2, 3, 1)$ and be parallel to the line $x + 3y - 2z - 2 = 0 = x - y + 2z$. If the distance of $L$ from the point $(5, 3, 8)$ is $\alpha$, then $3\alpha^{2}$ is equal to $\_\_\_\_$

Let the line $L: \frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{1}$ intersect the plane $2x + y + 3z = 16$ at the point $P$. Let the point $Q$ be the foot of perpendicular from the point $R(1,-1,-3)$ on the line $L$. If $\alpha$ is the area of triangle $PQR$, then $\alpha^2$ is equal to $\underline{\hspace{1cm}}$.

Let $\lambda _ { 1 } , \lambda _ { 2 }$ be the values of $\lambda$ for which the points $\left( \frac { 5 } { 2 } , 1 , \lambda \right)$ and $( - 2 , 0 , 1 )$ are at equal distance from the plane $2 x + 3 y - 6 z + 7 = 0$. If $\lambda _ { 1 } > \lambda _ { 2 }$, then the distance of the point $\left( \lambda _ { 1 } - \lambda _ { 2 } , \lambda _ { 2 } , \lambda _ { 1 } \right)$ from the line $\frac { x - 5 } { 1 } = \frac { y - 1 } { 2 } = \frac { z + 7 } { 2 }$ is $\_\_\_\_$

A line with direction ratio $2,1,2$ meets the lines $\mathrm { x } = \mathrm { y } + 2 = \mathrm { z }$ and $\mathrm { x } + 2 = 2 \mathrm { y } = 2 \mathrm { z }$ respectively at the point P and Q . if the length of the perpendicular from the point $( 1,2,12 )$ to the line PQ is $l$, then $l ^ { 2 }$ is

Let the position vectors of the vertices $A , B$ and $C$ of a triangle be $2 \hat { i } + 2 \hat { j } + \hat { k } , \hat { i } + 2 \hat { j } + 2 \hat { k }$ and $2 \hat { i } + \hat { j } + 2 \hat { k }$ respectively. Let $l _ { 1 } , l _ { 2 }$ and $l _ { 3 }$ be the lengths of perpendiculars drawn from the ortho centre of the triangle on the sides $A B , B C$ and $C A$ respectively, then $l _ { 1 } ^ { 2 } + l _ { 2 } ^ { 2 } + l _ { 3 } ^ { 2 }$ equals :
(1) $\frac { 1 } { 5 }$
(2) $\frac { 1 } { 2 }$
(3) $\frac { 1 } { 4 }$
(4) $\frac { 1 } { 3 }$

Let $\mathrm { P } ( \alpha , \beta , \gamma )$ be the image of the point $\mathrm { Q } ( 3 , - 3,1 )$ in the line $\frac { x - 0 } { 1 } = \frac { y - 3 } { 1 } = \frac { z - 1 } { - 1 }$ and R be the point $( 2,5 , - 1 )$. If the area of the triangle $PQR$ is $\lambda$ and $\lambda ^ { 2 } = 14 K$, then $K$ is equal to:
(1) 36
(2) 81
(3) 72
(4) 18

The square of the distance of the image of the point $( 6,1,5 )$ in the line $\frac { x - 1 } { 3 } = \frac { y } { 2 } = \frac { z - 2 } { 4 }$, from the origin is $\_\_\_\_$

Let a line pass through two distinct points $P ( - 2 , - 1,3 )$ and $Q$, and be parallel to the vector $3 \hat { i } + 2 \hat { j } + 2 \hat { k }$. If the distance of the point Q from the point $\mathrm { R } ( 1,3,3 )$ is 5 , then the square of the area of $\triangle P Q R$ is equal to :
(1) 148
(2) 136
(3) 144
(4) 140

The perpendicular distance, of the line $\frac { x - 1 } { 2 } = \frac { y + 2 } { - 1 } = \frac { z + 3 } { 2 }$ from the point $\mathrm { P } ( 2 , - 10,1 )$, is :
(1) 6
(2) $5 \sqrt { 2 }$
(3) $4 \sqrt { 3 }$
(4) $3 \sqrt { 5 }$

Let in a $\triangle ABC$, the length of the side $AC$ be 6, the vertex $B$ be $(1,2,3)$ and the vertices $A, C$ lie on the line $\frac{x-6}{3} = \frac{y-7}{2} = \frac{z-7}{-2}$. Then the area (in sq. units) of $\triangle ABC$ is:
(1) 17
(2) 21
(3) 56
(4) 42