(d) Potential due to charge $–q\ at\ A\ $
$V_{A} = \frac{1}{4\pi\varepsilon_{0}k}\frac{- q}{\left( r^{2} \right)^{1/2}}$
Potential due to charge $–q\ at\ B$
$V_{B} = \frac{1}{4\pi\varepsilon_{0}k}\frac{- q}{\left( r^{2} \right)^{1/2}}$
Potential due to charge $–q\ at\ C$
$V_{C} = \frac{1}{4\pi\varepsilon_{0}k}\frac{- q}{\left( r^{2} \right)^{1/2}}$
Total potential at centre.
$V = V_{A} + V_{B} + V_{C}$
$V = 0$
From charge configuration at the centre electric field in non zero.