If $\left(-2, 6\right)$ is the image of the point $\left(4, 2\right)$ with respect to the line $L=0,$ then $L=0$ is

Let $P\left(-2, 6\right) Q\left(4, 2\right)$

Slope of PQ $=\frac{2-6}{4+2}=\frac{-4}{6}=\frac{-2}{3}$

mid point of PQ $=\left(\frac{-2+4}{2}, \frac{6+2}{2}\right) = \left(1, 4\right)$

Slope of line $L= \frac{-1}{\left({\displaystyle \raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\right)}=\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

If P is Image of Q w.r.to $L=0$

Then line passes through (1, 4)

$\therefore$Eq of line $y-4=\frac{3}{2}\left(x-1\right)$

                     $$\begin{gathered} \Rightarrow 2y-8=3x-3 \\ \Rightarrow 3x-2y+5=0 \end{gathered}$$