Six point charges are kept at the vertices of a regular hexagon of side $L$ and centre $O$, as shown in the figure. Given that $K=\frac{1}{4\pi {\epsilon }_0}\frac{q}{L^2}$​, which of the following statement(s) is(are) correct.

(1) 

Resultant of $2K$ and $2K$( at $120°$) is also $2K$ towards $4K$. Therefore, net electric field is $6K$.

(2) $V_0=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q_A}{L}+\frac{q_B}{L}+\frac{q_C}{L}+\frac{q_D}{L}+\frac{q_E}{L}+\frac{q_F}{L}\right]$

$V_0=\frac{1}{4\pi {\epsilon }_{0}L}\left(q_A+...+q_F\right)=0$

Because $q_A+q_B+q_C+q_D+q_E+q_F=0$

(3) Only line $PR$, potential is same $\left(=0\right)$