Given equation of line is $x^{2}-4 x y+3 y^{2}=0$ $\therefore m_{1}+m_{2}=\frac{4}{3}$ and $m_{1} m_{2}=\frac{1}{3}$
On solving these equations, we get $m_{1}=1, m_{2}=\frac{1}{3}$
Let the lines parallel to given line are $y=m_{1} x+c_{1}$ and $y=m_{2} x+c_{2}$
$\therefore$
$y=\frac{1}{3} x+c_{1}$ and $y=x+c_{2}$
Also, these lines passes through the point $(3,-2)$
$\therefore$
$-2=\frac{1}{3} \times 3+c_{1}$
$\Rightarrow \quad c_{1}=-3$
and $\quad-2=1 \times 3+c_{2}$
$\Rightarrow \quad c_{2}=-5$
$\therefore$ Required equation of pair of lines is $(3 y-x+9)(y-x+5)=0$
$\Rightarrow x^{2}+3 y^{2}-4 x y-14 x+24 y+45=0$