If the line $\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2}$ and $\frac{7-7 x}{3…
If the line $\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2}$ and $\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles, then $p=$
$\frac{70}{11}$
$\frac{11}{70}$
$\frac{-70}{11}$
$\frac{-11}{70}$
Solution
Given lines can be written as
$\frac{x-1}{-3}=\frac{y-2}{\frac{2 p}{7}}=\frac{z-3}{2} \text { and } \frac{x-1}{3 p}=\frac{y-5}{1}=\frac{z-6}{-5}$
As these lines are at right angles, we get
$\begin{aligned}
& \quad-(-3)\left(-\frac{3 p}{7}\right)+\left(\frac{2 p}{7}\right)(1)+(2)(-5)=0 \\
& \therefore \quad \frac{9 p}{7}+\frac{2 p}{7}-10=0 \\
& \therefore \quad \frac{11 p}{7}=10 \Rightarrow p=\frac{70}{11}
\end{aligned}$