Explanation

(d)
Let $P(\alpha,\ \beta,\ \gamma)$ be the image of the point $Q( - 1,\ 3,4)$
Midpoint of $PQ$ lies on $x - 2y = 0$. Then,
$\frac{\alpha - 1}{2} - 2\left( \frac{\beta + 3}{2} \right) = 0$
$\Rightarrow \ \alpha - 1 - 2\beta - 6 = 0\ \Rightarrow \ \alpha - 2\beta = 7$ (i)
Also $PQ$ is perpendicular to the plane. Then,
$\frac{\alpha + 1}{1} = \frac{\beta - 3}{- 2} = \frac{\gamma - 4}{0}$ (ii)
Solving (i) and (ii), we get
$\alpha = \frac{9}{5},\beta = - \frac{13}{5},\gamma = 4$
Therefore, image is
$\left( \frac{9}{5}, - \frac{13}{5},4 \right)$
$Alternative\ method:$
For image,
$\frac{\alpha - ( - 1)}{1} = \frac{\beta - 3}{- 2} = \frac{\gamma - 4}{0} = \frac{- 2\left( - 1 - 2(3) \right)}{(1)^{2} + ( - 2)^{2}}$
$\Rightarrow \ \alpha = \frac{9}{5},\beta = - \frac{13}{5},\gamma = 4$