Let the perpendiculars from any point on the line $2x+11y=5$ upon the lines  $24x+7y-20=0$ and $4x-3y-2=0$ have the lengths $p_1$ and $p_2$ respectively. then, $p_1$-$p_2$ is equal to

Let $\left(t,\frac{5-2t}{11}\right)$ be a point on the line $2x+11y=5$.

Then,

$p_1=\left|\frac{24t+7\left(\frac{5-2t}{11}\right)-20}{\sqrt{{24}^2+7^2}}\right|=\frac{|50t-37|}{55}$

and $p_2=\left|\frac{4t-3\left(\begin{array}{c}5-2t\\ 11\end{array}\right)-2}{\sqrt{4^2+(-3{)}^2}}\right|=\frac{|50t-37|}{55}$

Clearly, we have $p_1$=$p_2$

$A\mathrm{LITER}$: Clearly, $2x+11y=5$  is the angle bisector of the two 

lines. Therefore $p_1$=$p_2$