(c)

When $q_1$ and $q_2$ are the magnitudes of charges at two vertices of an equilateral triangle of side $a$, magnitude of electric field at third vertex is

$E=\sqrt{\left(\frac{k q_1}{a}\right)^2+\left(\frac{k q_2}{a}\right)^2+\frac{2 k^2 q_1 q_2}{a^2} \cos 60^{\circ}}$

$=\frac{k}{a} \sqrt{q_1^2+q_2^2+q_1 q_2}$

$=\frac{k}{a} \sqrt{q_1^2+\left(q-q_1\right)^2+q_1\left(q-q_1\right)}$

$=\frac{k}{a q}\left(\sqrt{\left(\frac{q_1}{q}\right)^2+1-\left(\frac{q_1}{q}\right)}\right)$

$=\frac{k}{a q}\left(\sqrt{\left.\left(\frac{q_1}{q}-\frac{1}{2}\right)^2+\frac{3}{4}\right)}\right.$

So, field is minimum when $\frac{q_1}{q}=\frac{1}{2}$.

This condition is satisfied in graph $(c)$.