= Solution
Write $y_{i+}=\sum_jy_{ij}$ and $\mu_{i+}=\sum_j\mu_{ij}$. Under the model in part (a),
$$
\mu_{i+}=e^{\alpha_i}\sum_je^{\beta_j},
\qquad
p_j:=\frac{\mu_{ij}}{\mu_{i+}}
=\frac{e^{\beta_j}}{\sum_ke^{\beta_k}},
$$
and $p_j$ is independent of the price category $i$. The <Poisson-multinomial conditioning> identity gives
$$
(Y_{i1},Y_{i2},Y_{i3})\mid Y_{i+}=y_{i+}
\sim\operatorname{Multinomial}(y_{i+};p_1,p_2,p_3).
$$
More explicitly, the <likelihood function> for row $i$ factors as
$$
\prod_j\frac{e^{-\mu_{ij}}\mu_{ij}^{y_{ij}}}{y_{ij}!}
=
\frac{e^{-\mu_{i+}}\mu_{i+}^{y_{i+}}}{y_{i+}!}
\frac{y_{i+}!}{\prod_jy_{ij}!}\prod_jp_j^{y_{ij}}.
$$
The first factor is the likelihood of the row total and the second is the conditional <multinomial likelihood>. The free price effect $\alpha_i$ makes the first factor maximal at $\widehat\mu_{i+}=y_{i+}$, while maximizing the remaining factors gives
$$
\widehat p_j=\frac{y_{+j}}{y_{++}}.
$$
Consequently both fits have the same <maximum-likelihood fitted value>
$$
\boxed{\widehat\mu_{ij}=y_{i+}\widehat p_j
=\frac{y_{i+}y_{+j}}{y_{++}}}.
$$
Here every row total is $30$, the column totals are $31,28,31$, and $y_{++}=90$, so every fitted row is
$$
\boxed{(31/3,\ 28/3,\ 31/3)}.
$$
This factorization is the <Poisson trick>.
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