= Solution
The conditional output distributions are permutations of the same triple, so both <conditional entropy>[conditional entropies] equal
$$
H_0=H(1/6,1/3,1/2).
$$
With input probability $p$, the three output probabilities are $q_1=(2-p)/6$, $q_2=(1+2p)/6$, and $q_3=(3-p)/6$. Therefore
$$
\boxed{I(p)=-\sum_{j=1}^3q_j\log_2q_j-H_0}.
$$
Differentiating, the constant terms cancel because $\sum q_j'=0$:
$$
I'(p)=\frac16\log_2\frac{(2-p)(3-p)}{(1+2p)^2},\qquad
I''(p)=-\frac1{\log_e2}\sum_j\frac{(q_j')^2}{q_j}<0.
$$
Thus there is a unique maximum. Its stationary equation reduces to $3p^2+9p-5=0$, whose root in $[0,1]$ is
$$
\boxed{p_{\max}=\frac{\sqrt{141}-9}{6}}.
$$
The other root is negative; <strict concavity> proves this is the global maximum.
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