Suppose $\mathbf{Y} = \begin{bmatrix} Y_1 \\ Y_2 \\ Y_3 \end{bmatrix} \sim N(\mu, \Sigma)$ where $\mu = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ and $\Sigma = \begin{bmatrix} 2 & -1 & 1 \\ -1 & 5 & 1 \\ 1 & 1 & 3 \end{bmatrix}$. Calculate $\mathbb{E}[Y_3 | Y_1 = y_1, Y_2 = y_2]$ and $\mathbb{V}[Y_3 | Y_1 = y_1, Y_2 = y_2]$.
I know the general formula in the case when: $\mathbf{Y} = \begin{bmatrix} \mathbf{Y}_1 \\ \mathbf{Y}_2 \end{bmatrix} \sim N\left(\begin{bmatrix} \mu_1 \\ \mu_2 \end{bmatrix}, \begin{bmatrix} \Sigma_{11} & \Sigma_{12} \\ \Sigma_{21} & \Sigma_{22} \end{bmatrix} \right)$ then $\mathbf{Y}_2 | \mathbf{Y}_1 = \mathbf{y}_1 \sim N(\mu_2 + \Sigma_{21} \Sigma_{11}^{-1} ( \mathbf{y}_1 - \mu_1), \Sigma_{22} - \Sigma_{21} \Sigma_{11}^{-1} \Sigma_{12})$. However, I am unsure of how to apply it here, could someone please show me?
Partition the multivariate normal random vector $\mathbf{Y}$ consisting of two subvectors as
$\mathbf{Y}=\left(\begin{array}{c} \mathbf{Y^{(1)}}\\ \mathbf{Y^{(2)}}\\ \end{array} \right)$ where $\mathbf{Y^{(1)}}=\left(\begin{array}{c} Y_{1}\\ Y_{2}\\ \end{array} \right) $ and $\mathbf{Y^{(2)}}=\left(\begin{array}{c} Y_{3} \end{array} \right)$.
Accordingly partition the mean vector as $\mathbf{\mu}=\left(\begin{array}{c} \mathbf{\mu^{(1)}}\\ \mathbf{\mu^{(2)}}\\ \end{array} \right)$ where $\mathbf{\mu^{(1)}}=\left(\begin{array}{c} \mu_{1}\\ \mu_{2}\\ \end{array} \right) =\left(\begin{array}{c} 1\\ 2\\ \end{array} \right)$ and $\mathbf{\mu^{(2)}}=\left(\begin{array}{c} \mu_{3} \end{array} \right) = 3$.
and variance-covariance matrix as $\Sigma = \left(\begin{array}{ccc} 2 & -1 & 1\\ -1 & 5 & 1\\ 1 & 1 & 3\\ \end{array} \right) = \left(\begin{array}{cc|c} 2 & -1 & 1\\ -1 & 5 & 1\\ \hline 1 & 1 & 3\\ \end{array} \right) = \left(\begin{array}{cc} \Sigma_{11}&\Sigma_{12}\\ \Sigma_{21}&\Sigma_{22}\\ \end{array} \right)$
Plug-in the values in the expressions for mean and variance you have stated above to get the solution.