Let $P$ be the foot of the perpendicular from the point $Q ( 10 , - 3 , - 1 )$ on the line $\frac { x - 3 } { 7 } = \frac { y - 2 } { - 1 } = \frac { z + 1 } { - 2 }$. Then the area of the right angled triangle $P Q R$, where $R$ is the point $( 3 , - 2,1 )$, is
(1) $9 \sqrt { 15 }$
(2) $\sqrt { 30 }$
(3) $8 \sqrt { 15 }$
(4) $3 \sqrt { 30 }$

If the image of the point $( 4,4,3 )$ in the line $\frac { x - 1 } { 2 } = \frac { y - 2 } { 1 } = \frac { z - 1 } { 3 }$ is $( \alpha , \beta , \gamma )$, then $\alpha + \beta + \gamma$ is equal to
(1) 9
(2) 12
(3) 7
(4) 8

The square of the distance of the point $\left( \frac { 15 } { 7 } , \frac { 32 } { 7 } , 7 \right)$ from the line $\frac { x + 1 } { 3 } = \frac { y + 3 } { 5 } = \frac { z + 5 } { 7 }$ in the direction of the vector $\hat { i } + 4 \hat { j } + 7 \hat { k }$ is :
(1) 54
(2) 44
(3) 41
(4) 66

Let P be the foot of the perpendicular from the point $( 1,2,2 )$ on the line $\mathrm { L } : \frac { x - 1 } { 1 } = \frac { y + 1 } { - 1 } = \frac { z - 2 } { 2 }$. Let the line $\vec { r } = ( - \hat { i } + \hat { j } - 2 \hat { k } ) + \lambda ( \hat { i } - \hat { j } + \hat { k } ) , \lambda \in \mathbf { R }$, intersect the line L at Q . Then $2 ( \mathrm { PQ } ) ^ { 2 }$ is equal to :
(1) 25
(2) 19
(3) 29
(4) 27

In space, a plane E is given with points A and B on it, and a point P at a distance of 4 units from this plane.
The perpendicular projections of line segments PA and PB onto plane E, together with line segment AB, form an equilateral triangle with side length 2 units.
Accordingly, what is the product $| \mathbf { P A } | \cdot | \mathbf { P B } |$?
A) 8 B) 12 C) 16 D) 18 E) 20