Question

Find the values of $p$ so that the line $1-x$
$\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}$ and $\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles.

Answer 1

Consider the problem

Let

$\begin{array}{c}\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}\\ \\ \frac{-\left(x-1\right)}{3}=\frac{7\left(y-2\right)}{2p}=\frac{z-3}{2}\\ \\ \frac{x-1}{-3}=\frac{y-2}{\frac{2p}{7}}=\frac{z-3}{2}\end{array}$

Now,

$\begin{array}{c}\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}\\ \\ \frac{-7\left(x-1\right)}{3p}=\frac{y-5}{1}=\frac{-\left(z-6\right)}{5}\\ \\ \frac{x-1}{\frac{-3p}{7}}=\frac{y-5}{1}=\frac{\left(z-6\right)}{-5}\end{array}$

These two lines are perpendicular,

$-3\left(\frac{-3p}{7}\right)+1\left(\frac{2p}{7}\right)+2\left(-5\right)=0$

$\Rightarrow \frac{9p}{7}+\frac{2p}{7}-10=0$

$\Rightarrow \frac{11p}{7}=10$

$\Rightarrow p=\frac{70}{11}$