Let $(h, k)$ be any point on the line $7 x-9 y+10=0,$ then $7 h-9 k+10=0$
$\Rightarrow \quad 7 h=9 k-10$
$\Rightarrow \quad h=\frac{9 k-10}{7}$
Now, perpendicular distance from point
$(h, k)$ to the line $3 x+4 y=5$ is $d_{1}$
$d_{1}=\frac{3 h+4 k-5}{\sqrt{3^{2}+4^{2}}}$
$\Rightarrow \quad d_{1}=\frac{3 h+4 k-5}{5}$
and perpendicular distance from $(h, k)$ to the line $12 x+5 y=7$ is $\bar{d}_{2}$
$\therefore \quad d_{2}=\frac{12 h+5 k-7}{\sqrt{12^{2}+5^{2}}}$
$\Rightarrow \quad d_{2}=\frac{12 h+5 k-7}{13}$
Now, $d_{1}-d_{2}=\frac{3 h+4 k-5}{5}-\frac{12 h+5 k-7}{13}$
$\Rightarrow d_{1}-d_{2}$
$=\frac{13(3 h+4 k-5)-5(12 h+5 k-7)}{65}$
$=\frac{39 h+52 k-65-60 h-25 k+35}{65}$
$=\frac{-21 h+27 k-30}{65}$
$=\frac{-21\left(\frac{9 k-10}{7}\right)+27 k-30}{65} \quad[$ from Eq.(i) $]$
$=\frac{-27 k+30+27 k-30}{65}=0$
$\Rightarrow d_{1}-d_{2}=0 \Rightarrow d_{1}=d_{2}$