Find the image of the point $(2,–3,4)$ with respect to the plane  $4x+2y-4z+3=0$

The equation of a line passing through $(2,–3,4)$ and normal to a plane is

$\frac{x-2}{4}=\frac{y+3}{2}=\frac{z-4}{-4}$                            (i)

Any point on Eq. (i) is

$M(4\lambda +2,2\lambda -3,4-4\lambda )$

Since the point M lies on the plane, so

$\begin{array}{l}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}4(4\lambda +2)+2(2\lambda -3)-4(4-4\lambda )+3=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}36\lambda =11\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\lambda =\frac{11}{36}\end{array}$

Hence, the $\mathrm{M}$ point $=\left(\frac{29}{9},-\frac{33}{18},\frac{25}{9}\right)$

Let $Q(\alpha ,\beta ,\gamma )$ be the image of $\mathrm{P}(2,–3,4)$

be the image of $\mathrm{P}(2,–3,4)$

Thus,

 $\frac{\alpha +2}{2}=\frac{29}{9},\frac{\beta -3}{2}=\frac{-33}{18},\frac{\gamma +4}{2}=\frac{25}{9}$

$\Rightarrow (\alpha ,\beta ,\gamma )=\left(\frac{40}{9},-\frac{2}{3},\frac{14}{9}\right)$

which is the required image of $\mathrm{P}(2,–3,4)$.