Find \(p\) so that the lines given by \(\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}\) and \(\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}\) are perpendicular to each other

\(p=\frac{70}{11}\)

Given equation can be written in the standard form as

\(\frac{x − 1}{−2} = \frac{y − 2}{2p/7} = \frac{z − 3}{−5}\)

and \(\frac{x − 1}{−3p/7} = \frac{y −5}{1} = \frac{z − 6}{−5}\)

\(a_1 = −3, b_1 = 2p/7, c_1 = 2\)

\(a_2 = −3p/7, b_2 = 1, c_2 = −5\)

Two lines are perpendicular

\(⇒ a_1a_2 + b_1b_2 + c_1c_2 = 0\)

\(⇒ \frac{9p}{7} + \frac{2p}{7} − 10 = 0\)

\(⇒ p = \frac{70}{11}\)