= Solution
Let the input probabilities be $(s,t,u)$ and put $v=s+u=1-t$. The output distribution and <conditional entropy> are
$$
q=\left(\frac s2,1-\frac v2,\frac u2\right),\qquad H(Y\mid X)=s+u=v.
$$
For fixed $v$, the first and third output masses <sum> to $v/2$. Their <Shannon entropy>[entropy] is maximized when they are equal, so $s=u=v/2$. Equivalently, if $v>0$,
$$
H(Y)=h_2(v/2)+(v/2)h_2(s/v)\leq h_2(v/2)+v/2.
$$
The same bound holds at $v=0$ by <continuity>. Therefore $I(X;Y)\leq h_2(v/2)-v/2$, with equality for symmetric outer input probabilities. Writing $z=v/2\in[0,1/2]$, differentiate:
$$
\frac d{dz}(h_2(z)-z)=\log_2\frac{1-z}z-1.
$$
The unique zero is $z=1/3$; strict <concavity> makes it the global maximum. Thus $v=2/3$ and
$$
\boxed{p_X=(1/3,1/3,1/3),\qquad C=h_2(1/3)-1/3=\log_2(3/2).}
$$
The optimum output is $(1/6,2/3,1/6)$.
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