= Solution
The stationary equations simplify by setting $a_i=\pi_i\lambda_i$. They read $2a_i=\sum_{j\ne i}a_j$, so all three $a_i$ are equal. If all rates are positive, this gives the unique equilibrium
$$
\boxed{\pi_i=\frac{\lambda_i^{-1}}{\lambda_1^{-1}+\lambda_2^{-1}+\lambda_3^{-1}}.}
$$
If any rate is zero, the common $a_i$ is zero, forcing zero equilibrium mass on every positive-rate state. Every zero-rate state is absorbing. With one such state $z$, the unique equilibrium is $\delta_z$; with two, all probability mixtures on those two states are equilibria, giving nonuniqueness. The excluded all-zero case would make every distribution an equilibrium.
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