Step 1 :
Equation of the given line is $5x-12y-3$.
Let the point be $(x_1, y_1)$
Hence the distance of this line from the point $(x_1, y_1)$ is
$ d = \bigg| \large\frac{5x_1-12y_1-3}{\sqrt{5^2+12^2}} \bigg|$$ = \large\frac{5x_1-12y_1-3}{13}$
Also the distance between the points $(x_1, y_1)$ and (3, -2) is
$ d = \sqrt{(x_1-3)^2+(y_1+2)^2}$
According to the given conditions
$( \sqrt{(x_1-3)^2+(y_1+2)^2} )^2$ = $ \large\frac{(5x_1-12y_1-3)}{13} $
(i.e) $x_1^2-6x_1+9+y_1^2+4y_1+4 = \large\frac{(5x-12y-3)}{13}$
$ \Rightarrow 13(x_1^2-6x_1+9 + y_1^2+4y_1+4)= 5x-12y-3$
$ \Rightarrow 13(x_1^2+y_1^2)-83x_1+64y_1+172=0$
Hence the equation of the locus of the point is $13(x^2+y^2)-83x+64y+172=0$