Partition with $a=(1,2)$ and $b=3$ :

$\mu_a=\begin{pmatrix}1\\0\end{pmatrix},\quad \mu_b=3,\quad \Sigma_{aa}=\begin{pmatrix}2&1\\1&4\end{pmatrix},\quad \Sigma_{ab}=\begin{pmatrix}0.5\\1\end{pmatrix},\quad \Sigma_{bb}=1 .$

Conditional mean.

$\mu_{a\mid b}=\mu_a+\Sigma_{ab}\Sigma_{bb}^{-1}(x_b-\mu_b) =\begin{pmatrix}1\\0\end{pmatrix} +\begin{pmatrix}0.5\\1\end{pmatrix}(2-3) =\begin{pmatrix}1-0.5\\0-1\end{pmatrix} =\begin{pmatrix}0.5\\-1\end{pmatrix}.$

The observed $X_3=2$ is below its mean of $3$ , and both covariances with $X_3$ are positive, so both conditional means shift down. That sign check alone eliminates (A) and (B).

Conditional covariance.

$\Sigma_{a\mid b}=\Sigma_{aa}-\Sigma_{ab}\Sigma_{bb}^{-1}\Sigma_{ba} =\begin{pmatrix}2&1\\1&4\end{pmatrix} -\begin{pmatrix}0.5\\1\end{pmatrix}\begin{pmatrix}0.5&1\end{pmatrix} =\begin{pmatrix}2&1\\1&4\end{pmatrix} -\begin{pmatrix}0.25&0.5\\0.5&1\end{pmatrix} =\begin{pmatrix}1.75&0.5\\0.5&3\end{pmatrix}.$

Conditioning reduces every variance, as it must — information can only sharpen. Note also that it does not depend on the value $2$ .