Explanation

(b)
Given plane is $\overrightarrow{r} \bullet \overrightarrow{n} = q$ (i)
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Let the image of $A(\overrightarrow{a})$ in the plane be $B(\overrightarrow{b})$
Equation of $AC$ is $\overrightarrow{r} = \overrightarrow{a} + \lambda\overrightarrow{n}$ ($\because AC$ is normal to the plane) (ii)
Solving (i) and (ii), we get
$\left( \overrightarrow{a} + \lambda\overrightarrow{n} \right).\overrightarrow{n} = q$
$\Rightarrow \ \lambda = \frac{q - \overrightarrow{a} \bullet \overrightarrow{n}}{\left| \overrightarrow{n} \right|^{2}}$
$\therefore\overrightarrow{OC} = \overrightarrow{a} + \frac{(q - \overrightarrow{a} \bullet \overrightarrow{n})}{\left| \overrightarrow{n} \right|^{2}} \bullet \overrightarrow{n}$
But $\overrightarrow{OC} = \frac{\overrightarrow{a} + \overrightarrow{b}}{2}$
$\therefore\ \overrightarrow{a} + \frac{(q - \overrightarrow{a} \bullet \overrightarrow{n})\overrightarrow{n}}{\left| \overrightarrow{n} \right|^{2}} = \frac{\overrightarrow{a} + \overrightarrow{b}}{2}$
$\Rightarrow \overrightarrow{b} = \overrightarrow{a} + 2\left( \frac{q - \overrightarrow{a} \bullet \overrightarrow{n}}{\left| \overrightarrow{n} \right|^{2}} \right)\overrightarrow{n}$