The pair of straight line
$$
\begin{array}{rl}
\Rightarrow 2 x^2+(-y+6) x+\left(-3 y^2+y+4\right)=0 & 2 x^2-x y-3 y^2+6 x+y+4=0 \\
\therefore \quad x & =\frac{(y-6) \pm \sqrt{(y-6)^6-4 \times 2\left(-3 y^2+y+4\right)}}{2 \times 2} \\
& =\frac{(y-6) \pm \sqrt{y^2+36-12 y+24 y^2-8 y-32}}{4} \\
& =\frac{(y-6) \pm \sqrt{25 y^2+4-20 y}}{4} \\
& =\frac{(y-6) \pm \sqrt{(5 y-2)^2}}{4} \\
x & =\frac{(y-6) \pm(5 y-2)}{4} \\
\text { So, } x & =\frac{y-6+(5 y-2)}{4} \text { and } x=\frac{y-6-5 y+2}{4} \\
\Rightarrow 4 x & =6 y-8 \text { and } 4 x=-4 y-4 \\
\Rightarrow 2 x & -3 y+4=0 \text { and } x+y+1=0
\end{array}
$$
Now, product length of perpendicular from $(-1,5)$ to both lines are
$\begin{aligned} & P_1 \cdot P_2=\left|\frac{-2-3(5)+4}{\sqrt{2^2+(-3)^2}}\right|\left|\frac{-1+5+1}{\sqrt{1^2+1^2}}\right| \\ & =\frac{13}{\sqrt{13}} \times \frac{5}{\sqrt{2}}=\frac{65}{\sqrt{26}}\end{aligned}$