The reflection of the point $\overrightarrow{\mathrm{a}}$ in the plane $\overrightarrow{\mathrm{r}}\cdot \overrightarrow{\mathrm{n}}=\mathrm{q}$ is

Given plane is  $\overrightarrow{\mathrm{r}}\cdot \overrightarrow{\mathrm{n}}=\mathrm{q}---\left(\mathrm{i}\right)$

Let the image of $\mathrm{A}\left(\overrightarrow{\mathrm{a}}\right)$ in the plane be $\mathrm{B}\left(\overrightarrow{\mathrm{b}}\right)$

Equation of $\mathrm{AC}\text{ is }\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\mathrm{\lambda }\overrightarrow{\mathrm{n}}$

$(\because AC\text{ is normal to the plane) ----}\left(ii\right)$

Solving (i) and (ii), we get 

$\begin{array}{l}(\overrightarrow{\mathrm{a}}+\mathrm{\lambda }\overrightarrow{\mathrm{n}})\cdot \overrightarrow{\mathrm{n}}=\mathrm{q}\\ \mathrm{or} \mathrm{\lambda }=\frac{\mathrm{q}-\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{n}}}{|\overrightarrow{\mathrm{n}}{|}^2}\\ \therefore \overrightarrow{\mathrm{OC}}=\overrightarrow{\mathrm{a}}+\frac{(\mathrm{q}-\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{n}})}{{\left.\overrightarrow{\mathrm{a}}\right|}^2}\cdot \overrightarrow{\mathrm{n}}\\ \mathrm{but} \overrightarrow{\mathrm{OC}}=\frac{\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}}{2}\\ \therefore \overrightarrow{\mathrm{a}}+\frac{(\mathrm{q}-\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{n}})\overrightarrow{\mathrm{n}}}{|\overrightarrow{\mathrm{n}}{|}^2}=\frac{\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}}{2}\\ \mathrm{or} \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{a}}+2\left(\frac{\mathrm{q}-\overrightarrow{\mathrm{a}}\cdot \overrightarrow{\mathrm{n}}}{|\overrightarrow{\mathrm{n}}{|}^2}\right)\overrightarrow{\mathrm{n}}\end{array}$