 Solution

Take the electrostatic potential energy to vanish for infinitely separated point charges, and write $k_e=1/(4\pi {ϵ}_0)$. The first charge costs no work to place. Bringing the second to distance $a$ costs $k_e{q}^2/a$; the third is distance $a$ from both existing charges and costs $2k_e{q}^2/a$. Thus the external quasistatic assembly work isEquivalently, sum over the three distinct pairs in the equilateral triangle. Counting both ordered pairs without a compensating factor of one half would double the answer. The positive result applies to either sign of $q$, since all three electric charges have the same sign.