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 = 1/(4πε₀qL²), which of the following statement(s) is(are) correct?

The distances from the centre to each charge are equal, i.e., r = L/(2sin30°) = L. The field at O due to charge (+2q) at A is |→EA| = 1/(4πε₀)(2q/r²) = 1/(4πε₀)(2q/L²) = 2K. Similarly, |→ED| = 2K, |→EB| = K, |→EC| = K, |→EE| = K, |→EF| = K. The field at O is EO = EA + ED + (EB + EE)cos60° + (EC + EF)cos60° = 2K + 2K + 2K(1/2) + 2K(1/2) = 6K. Potential at O is VO = 1/(4πε₀r)[2q + q - q - 2q - q + q] = 0. As line PR is perpendicular bisector for the electric dipoles AD, FE, and BC, it is an equatorial line and potential at all points on PR will be the same. At any point ST, the electric field will be directed from S to T. The potential decreases along the electric field line.