The values of p for which the lines $\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2}$ and $\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles is:

$\frac{70}{11}$

The correct answer is Option (1) → $\frac{70}{11}$

Rewriting equation of lines in standard form

line 1: $\frac{x-1}{-3}=\frac{y-2}{2p/7}=\frac{z-3}{2}$

line 2: $\frac{x-1}{-3p/7}=\frac{y-5}{1}=\frac{z-6}{-5}$

$\vec{v_1}$ || line 1, $\vec{v_2}$ || line 2

$\vec{v_1}=-3\hat i+\frac{2p}{7}\hat j+2\hat k$

$\vec{v_2}=\frac{-3p}{7}\hat i+\hat j-5\hat k$

for them to be $\vec{v_1}⊥\vec{v_2}$

$\vec{v_1}.\vec{v_2}⇒\frac{9p}{7}+\frac{2p}{7}-10=0⇒11p=70$

$⇒p=\frac{70}{11}$