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

The Correct Option is C

The correct answer is option (C): \(\frac{32}{3}\)
a(x - 3)+b(y + 4)+c(z - 7) = 0
P :   9a - b -5c = 0
    -11a - b + 5c = 0
After solving DR's ∝ (1, -1 , 2)
Equation of plane
x - y + 2z = 21
\(d=\frac{8}{\sqrt6}\)
\(d^2=\frac{32}{2}\)