Question

Let $P$ be the plane containing the straight line $\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}$ and perpendicular to the plane containing the straight lines $\frac{ x }{2}=\frac{ y }{3}=\frac{ z }{5}$ and $\frac{ x }{3}=\frac{ y }{7}=\frac{ z }{8}$. If $d$ is the distance of $P$ from the point $(2,-5,11)$, then $d ^2$ is equal to :

• A. $\frac{147}{2}$
• B. 96
• C. $\frac{32}{3}$
• D. 54

Answer 1

Answer: (C)

$a ( x -3)+ b ( y +4)+ c ( z -7)=0$
$ P : 9 a - b -5 c =0$
$ -11 a - b +5 c =0$
After solving DR's $\propto(1,-1,2)$
Equation of plane
$x - y +2 z =21 $
$ d =\frac{8}{\sqrt{6}} $
$ d ^2=\frac{32}{3}$