= Solution
First realize the chain as a <random walk on a multigraph>: put three parallel edges between $a,b$, two between $a,c$, and one between $b,c$. There are six edges and vertex degrees $5,4,3$. Choosing an incident edge uniformly gives exactly the stated transition probabilities. The <stationary distribution of a graph random walk> is the degree divided by twelve, $\pi=(5,4,3)/12$, and detailed balance follows because each edge multiplicity contributes the same flux in both directions.
For the first positive return time, <Kac's lemma> gives
$$
\boxed{k(a,a)=1/\pi_a=12/5.}
$$
For passage to $b$, put $h_a=k(a,b)$ and $h_c=k(c,b)$. The first-step equations are $h_a=1+(2/5)h_c$ and $h_c=1+(2/3)h_a$, since reaching $b$ stops the clock. Thus
$$
\boxed{k(a,b)=21/11,\qquad k(c,b)=25/11.}
$$
A multigraph is essential: a simple graph on three vertices cannot have six edges.
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