Let $\rho=a/\sqrt3$ be the distance from the origin to each fixed charge, and denote their position vectors by $\mathbf R_j$. For a small planar displacement $\mathbf r$, the <Taylor expansion>
$$
\frac1{|\mathbf R_j-\mathbf r|}=\frac1\rho+\frac{\mathbf R_j\cdot\mathbf r}{\rho^3}+\frac{3(\mathbf R_j\cdot\mathbf r)^2-\rho^2|\mathbf r|^2}{2\rho^5}+O(|\mathbf r|^3)
$$
can be summed using the threefold <symmetry> relations
$$
\sum_j\mathbf R_j=0,
\qquad
\sum_j\mathbf R_j\mathbf R_j^{\mathsf T}=\frac{3\rho^2}{2}I.
$$
The <electric potential> due to the fixed charges is consequently
$$
\Phi(\mathbf r)=\frac{q}{4\pi\epsilon_0}\left(\frac3\rho+\frac{3|\mathbf r|^2}{4\rho^3}+O(|\mathbf r|^3)\right).
$$
The moving charge has <electrostatic potential energy of point charges> $U=q\Phi$. Because the quadratic coefficient is positive, the origin is a strict <local minimum> of $U$, and the equilibrium is stable.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 4 / 10A / b / ii / Solution

For distinct <point charges> $q_i$ at positions $r_i$, the electrostatic <potential energy> is
$$
U=\frac1{4\pi\epsilon_0}\sum_{i<j}\frac{q_iq_j}{|r_i-r_j|}.
$$
When some charges are fixed, the terms depending on a movable charge form its potential energy in their <electric potential>.

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 Incoming links: Electrostatic potential energy of point charges