= Solution
The function $U(s)=V_0e^{-2s}$ is strictly <convex function>[convex]. Subject to $\alpha+\beta+\gamma=2\pi$, the <Jensen inequality> gives
$$
U(\alpha)+U(\beta)+U(\gamma)
\geq3U\left(\frac{\alpha+\beta+\gamma}{3}\right),
$$
with equality only at
$$
\boxed{\alpha=\beta=\gamma=\frac{2\pi}{3}}.
$$
This is the minimum of the relative potential, so it is a <stable equilibrium in Lagrangian mechanics>[stable relative equilibrium].
Let $q_j=q_j^{(0)}+x_j$ near an equally spaced configuration, and put
$$
A=V_0e^{-4\pi/3}.
$$
The gap perturbations are
$$
\delta\alpha=x_2-x_1,\qquad
\delta\beta=x_3-x_2,\qquad
\delta\gamma=x_1-x_3,
$$
whose sum vanishes. A <Taylor series>[Taylor expansion] gives the quadratic energies
$$
T_2=\frac{mr^2}{2}\sum_j\dot x_j^2,
\qquad
V_2=2A\left[
(x_2-x_1)^2+(x_3-x_2)^2+(x_1-x_3)^2
\right].
$$
Thus the linearized equations are
$$
mr^2\ddot x_i+4A(2x_i-x_j-x_k)=0,
\qquad \{i,j,k\}=\{1,2,3\}.
$$
The matrix in parentheses is the <Graph Laplacian> of $K_3$. Its constant eigenvector has eigenvalue zero, while every vector whose components sum to zero has eigenvalue three. One orthonormal set of three <normal modes> and their <angular frequencies> is therefore
$$
\begin{array}{c|c}
\text{mode}&\omega\\ \hline
\displaystyle \frac1{\sqrt3}(1,1,1)&0\\[2mm]
\displaystyle \frac1{\sqrt2}(1,-1,0)&
\displaystyle\sqrt{\frac{12V_0e^{-4\pi/3}}{mr^2}}\\[3mm]
\displaystyle \frac1{\sqrt6}(1,1,-2)&
\displaystyle\sqrt{\frac{12V_0e^{-4\pi/3}}{mr^2}}
\end{array}
$$
The zero mode is rigid rotation. The positive and equal squared frequencies on the two-dimensional relative subspace prove stability and agree with the general <normal modes of three equal masses with a symmetric gap potential>.
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