= Solution
Put $D^*=D-A$. By (i), $D^*X=0$. Thus $D^*A^T=D^*X(X^TX)^{-1}=0$, and its transpose vanishes too. Expanding $DD^T=(D^*+A)(D^*+A)^T$ yields
$$
\boxed{\operatorname{Var}(\tilde\theta)=\sigma^2\left(D^*D^{*T}+(X^TX)^{-1}\right).}
$$
The <covariance> difference is <positive semidefinite>. In particular its $i$th diagonal entry is $\sigma^2\sum_j(D^*_{ij})^2\ge0$, proving the requested componentwise <variance> inequalities. Prefer $\boxed{\hat\theta_1}$ within the class of linear unbiased estimators, since it has the least <variance> and the same <expectation>. A tie is possible if the first row of $D^*$ is zero. This is the <Gauss-Markov theorem> conclusion; no normal-distribution assumption on the errors is required.
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