Joint equation of pair of lines through \((3,-2)\) and parallel to \(x^{2}-4 x y+3 y^{2}=0\) is

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 \((3,-2)\)
\(\therefore\) \(-2=\frac{1}{3} \times 3+c_{1}\)
\(\Rightarrow c_{1}=-3\)
and \(\quad-2=1 \times 3+c_{2}\)
\(\Rightarrow c_{2}=-5\)
\(\therefore\) Required equation of pair of lines is \((3 y-x+9)(y-x+5)=0\)
\(\Rightarrow x^{2}+3 y^{2}-4 x y-14 x+24 y+45=0\)