= Solution
Write the partitioned mean as $(\mu_1^T,\mu_2^T)^T$ and assume $V_{22}$ is nonsingular, as required by the displayed inverse. For the <conditional multivariate normal distribution>, form the residual
$$
U=X_1-\mu_1-V_{12}V_{22}^{-1}(X_2-\mu_2).
$$
A <linear transformation> of a <multivariate normal distribution> is again normal. Direct calculation gives
$$
\operatorname{Cov}(U,X_2)=V_{12}-V_{12}V_{22}^{-1}V_{22}=0,\qquad \operatorname{Cov}(U)=V_{11}-V_{12}V_{22}^{-1}V_{21}.
$$
The jointly normal blocks $U$ and $X_2$ are therefore <independent>: their joint <characteristic function> has no cross term in its quadratic exponent and factors into the two marginal <characteristic functions>. Conditioning leaves the distribution of $U$ unchanged. Reconstructing $X_1$ proves the full conditional law, including the requested <Schur complement> <covariance>:
$$
\boxed{X_1\mid X_2=x_2\sim N\!\left(\mu_1+V_{12}V_{22}^{-1}(x_2-\mu_2),\ V_{11}-V_{12}V_{22}^{-1}V_{21}\right).}
$$
This describes a regular conditional law; it does not require the equality event to have positive probability.
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