To calculate the conditional distribution of multivariate normal random vector $(X,Y)$, since that for all $a \in \mathbb{R}$ the random vector $(X-aY,Y)$ is normally distributed, we can determine $a$ in such way that $X-aY$ and $Y$ are not correlated (so, in this case, independent):
$$0 = \mathrm{Cov}(X-aY,Y)= \mathrm{Cov}(X,Y)-a\mathrm{Var}(Y)$$ $$a=\frac{\mathrm{Cov}(X,Y)}{\mathrm{Var}(X)}$$
Then, with that $a$, we can write $X = X-aY+aY = Z+aY$ where $Z=X-aY$ and $Y$ are independent, hence the conditional distribution of $X|Y=y$ is the distribution of the random variable $Z+ay$, intuitively it's clear, but how to prove it rigorously?
Once this is done, we know that $Z+ay$ is normal multivariate and:
$$\mathrm{Var}(Z+ay)=\mathrm{Var}(X)+a^2\mathrm{Var}(Y)-2a\mathrm{Cov}(X,Y)=\mathrm{Var}(X)-\frac{\mathrm{Cov}(X,Y)^2}{\mathrm{Var}(Y)}$$
$$\mathrm{E}(Z+ay)=\mathrm{E}(X)+\frac{\mathrm{Cov}(X,Y)}{\mathrm{Var}(Y)}(y-\mathrm{E}(Y))$$