Six point charges are kept at the vertices of a regular hexagon of...

Here $$\begin{aligned} & \frac{\left|\overrightarrow{E_{A}}\right|}{2}=\left|\overrightarrow{E_{B}}\right|=\left|\overrightarrow{E_{C}}\right|=\frac{\left|\overrightarrow{E_{D}}\right|}{2}=\left|\overrightarrow{E_{E}}\right|=\left|\overrightarrow{E_{F}}\right|=K \\ \therefore & E_{O}=E_{A}+E_{D}+\left(E_{F}+E_{C}\right) \cos 60^{\circ}+\left(E_{B}+E_{C}\right) \cos 60^{\circ} \\ =& 2 \mathrm{~K}+2 \mathrm{~K}+(\mathrm{K}+\mathrm{K}) \times \frac{1}{2}+(\mathrm{K}+\mathrm{K}) \times \frac{1}{2}=6 \mathrm{~K} \end{aligned}$$
Electric potential at $O V_{O}=\frac{1}{4 \pi \varepsilon_{0} L}[2 q+q+q-q-q-2 q]=0$ Potential at all points on the line PR is same not on line ST. $P R$ is perpendicular bisector (the equatorial line) for the electric dipoles $A B, F E and B C$. Therefore the electric potential will be zero at any point on $P R$.