Let P be the plane containing the straight line {x-3}/9 = {y+4}/-1 = {z-7}/-5 and perpendicular to the plane ​

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 \(\frac x2 = \frac y3 = \frac z5\) and \(\frac x3 = \frac y7 = \frac z8\). If d is the distance of P from the point (2, –5, 11), then d² is equal to : 

(A) \(\frac{147}2\)

(B) 96

(C) \(\frac{32}3\)

(D) 54

Answer 1

Correct option is (C) \(\frac{32}3\)