= Solution
For the oriented edge $i\to j=i+1\pmod3$, $J_i=\alpha P_i-\beta P_j$. Substitution in the Markov-chain entropy production gives
$$
\dot S_{\rm tot}=\sum_iJ_i\log\frac{\alpha P_i}{\beta P_j}.
$$
Since $\sum_iJ_i=\alpha-\beta$,
$$
\boxed{\dot S_{\rm tot}=(\alpha-\beta)\log\frac\alpha\beta
+\sum_i[\alpha P_i-\beta P_j]\log\frac{P_i}{P_j}.}
$$
The second term is relaxational and vanishes for the uniform stationary distribution. The steady entropy exported to the environment is therefore
$$
\boxed{\dot S_{\rm neq}=(\alpha-\beta)\log(\alpha/\beta)\geq0.}
$$
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