Let the required point be $\mathrm{P}\left(x_1, y_1\right)$ and it is on the line $4 x-y-2=0$. Then,


Also given, point $P$ is equidistant from $A(-5,6)$ and
$\begin{aligned}
& B(3,2) . \\
& \therefore \quad P A^2=P B^2 \\
& \left(x_1+5\right)^2+\left(y_1-6\right)^2=\left(x_1-3\right)^2+\left(y_1-2\right)^2 \\
& x_1^2+10 x_1+25+y_1^2-12 y_1+36 \\
& =x_1^2-6 x_1+9+y_1^2-4 y_1+4 \\
& =16 x_1-8 y_1+48=0
\end{aligned}$

On solving Eq. (i) in Eq. (ii), we get
$\begin{aligned}
& \therefore \quad x_1=4 \\
& \text { and } y_1=14
\end{aligned}$