The equation of the lines given are

$ \qquad x+y-5=0$--------(1)

$ \qquad 3x-2y+7=0$------(2)

The perpendicular distance of line (1) from the point $p(x,y) is

$ d_1 = \bigg| \large\frac{x+y-5}{\sqrt{1^2+1^2}} \bigg|$

$ = \large\frac{|x+y-5|}{\sqrt 2}$

Similarly the perpendicular distance of line (2) from point $p(x,y)$ is

$ d_2 = \bigg| \large\frac{3x-2y+7}{\sqrt{(3)^2+(-2)^2}} \bigg|$

$ = \large\frac{|3x-2y+7|}{\sqrt{13}}$

It is given that sum of the perpendicular distance is 10.

(i.e.,) $d_1+d_2=10$

(i.e., )$ \large\frac{|x+y-5|}{\sqrt 2}$$+ \large\frac{|3x-2y+7|}{\sqrt{13}}$$ = 10$

(i.e., ) $ \sqrt{13}\: |x+y-5|+\sqrt 2 |3x-2y+7| = 10 \sqrt{26}$

Let us take the positive values of $(x+y-5)$ and $(3x-2y+7)$

$ \sqrt{13} (x+y-5)+ \sqrt 2 (3x-2y+7)-10 \sqrt{26}=0$

$ \Rightarrow x( \sqrt{13}+3\sqrt 2)+y ( \sqrt{13}-2\sqrt 2)+ ( 7\sqrt 2-5 \sqrt {13}-10\sqrt{26}) = 0$

This represents the equation of a straight line.

This proves that the point $p$ must move on a line.