Since, Eq. (i) passes through $(10,-7)$ which is point of intersection of given lines.

Also given, slope $=\frac{2}{3} \Rightarrow \frac{-\frac{1}{a}}{\frac{1}{b}}=\frac{2}{3}$
$\Rightarrow \quad b=\frac{-2 a}{3} \quad[$ Put in Eq. (ii) $] \ldots$ (iii)
$\begin{aligned} & \frac{10}{a}+\frac{7}{\frac{2 a}{3}} =1 \Rightarrow a=\frac{41}{2} \\ \therefore & b =-\frac{2 a}{3}=\frac{-2 \times \frac{41}{2}}{3}=-\frac{41}{3} \\ \therefore & a+b =\frac{41}{2}-\frac{41}{3}=\frac{123-82}{6}=\frac{41}{6}\end{aligned}$
Since, Eq. (i) passes through $\left(0, \frac{4}{5}\right)$, then
$$
2 \times 0-5 \times \frac{4}{5}+\mu=0 \Rightarrow \mu=4
$$
From Eq. (i),
$$
2 x-5 y+4=0
$$