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)

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