= Solution
The two design matrices span the same six-dimensional space of arbitrary log means for the six periods. Fit.2 uses the first period as a baseline, whereas fit.3 uses cumulative indicator steps. If $a_j$ is the log mean for period $j$, fit.3 has intercept $a_1$ and jump coefficients $a_{j+1}-a_j$. Thus both fits give identical fitted means and identical maximized <likelihood functions>. Therefore
$$
\boxed{D(\text{fit.2})-D(\text{fit.3})=0.}
$$
Different coefficient tests arise because they test different contrasts, not because one full fit explains the observations better.
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