Explanation

(c) In an equilateral triangle distance of centroid from all the vertices is same (say$r$).
$\therefore\ \ V = V_{1} + V_{2} + V_{3} = \frac{1}{4\pi\varepsilon_{0}}\left\lbrack \frac{2q}{r} - \frac{q}{r} - \frac{q}{r} \right\rbrack = 0$
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But ${\overrightarrow{E}}_{A}$=$\frac{1}{4\pi\varepsilon_{0}}.\frac{2q}{1}\ along\ AO,\ {\overrightarrow{E}}_{B} = \frac{1}{4\pi\varepsilon_{0}}\frac{q}{r^{2}}$ along $OB$ and
${\overrightarrow{E}}_{c} = \frac{1}{4\pi\varepsilon_{0}}.\frac{q}{r^{2}}\ along\ OC.obviously\ {\overrightarrow{E}}_{B} + {\overrightarrow{E}}_{B}\ $Will also be in the direction of $AO$ extended and hence $\ {\overrightarrow{E}}_{A}\ and\left( {\overrightarrow{E}}_{B} + {\overrightarrow{E}}_{C} \right)$ being in same direction will not give zero resultant.