Let $(X,Y,Z)$ a normal trivariate distribution, with expectation $0$, i.e. $$(\mathbb E(X), \mathbb E(Y), \mathbb E(Z)) = (0,0,0)$$ and matrix of variances-covariances $$\begin{pmatrix} 1 & 0 & \alpha \\ 0 & 1 & \beta \\ \alpha & \beta & 1 \end{pmatrix}$$
It's easy to see that $$\mathbb E(Y|X) = 0 \hspace{1cm} \text{and} \hspace{1cm} \mathbb E(Y|X,Z) = \frac{-\alpha \beta}{1-\beta^2}X + \frac{\alpha}{1-\beta^2}Z.$$
Unfortunately I'm blind and unable to see that! I see that $\mathbb E(XY)=0$ and $\mathbb E(XZ)= \alpha$ and $\mathbb E(Y,Z) = \beta$.
Thanks!
When $(X,Y,Z)$ are jointly normal, the conditional expectation of $Y$ given $X, Z$ is a linear combination of $X$ and $Z$, expressible in matrix algebra as: $$ E(Y\mid X,Z) = E(Y)+\left[ \operatorname{cov}(Y,X)\ \ \operatorname{cov}(Y,Z)\right]\left[ \begin{matrix} \operatorname{var}(X) & \operatorname{cov}(X,Z) \\ \operatorname{cov}(Z,X) & \operatorname{var}(Z) \end{matrix}\right]^{-1}\left[\begin{matrix} X-EX \\ Z-EZ \end{matrix}\right] $$ Plug in the particulars of your problem and grind it through. The general result can be found here.