Explanation

(c) The potential due to charge q at a distance $r$ is given by
$V = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{r}$
Since, potential is a scalar quantity, it can be added to find the sum due to individual charges.
$\sum V = V_{A} + V_{B} + V_{C}$
$V_{A} = \frac{1}{4\pi\varepsilon_{0}}\ .\frac{2q}{x}$
$V_{B} = - \frac{1}{4\pi\varepsilon_{0}}.\frac{q}{x}$
$V_{C} = - \frac{1}{4\pi\varepsilon_{0}}.\frac{q}{x}\ $
$\therefore\ \ V = \frac{1}{4\pi\varepsilon_{0}}\left( \frac{2q}{x} - \frac{q}{x} - \frac{q}{x} \right) = 0\ $
Electric field is a vector quantity, hence component along $OD$ is taken
$E = \frac{1}{4\pi\varepsilon_{0}}\left( \frac{2q}{x^{2}} + \frac{2q}{x^{2}}\cos\theta \right) \neq 0$