How to find conditional expectation $E(X|Y,Z)$? Question:
Random variable $X$, $Y$ and $Z$ are as follows:
$$Y=X+a$$
$$Z=X+b$$
where
$$a\sim \text{normal}(0, \sigma_y^2)$$
and 
$$b\sim \text{normal}(0, \sigma_z^2)$$
$a$, $b$ and $X$ are independent of each other.
Then how to find $E(X|Y=y, Z=z)$?
 A: Let $X, A, B$ denote independent normal random variables such that $E[A]=E[B]=0$, and let $Y=X+A$, and $Z=X+B$. Then, $X, Y$, and $Z$ are jointly normal random variables with the same mean $\mu = E[X]$. Given $Y=y$ and $Z=z$, $X$ is a (conditionally) normal random variable with conditional mean of the form $\alpha y + (1-\alpha)z$ where $\alpha \in (0,1)$, that is, 
$$E[X\mid Y, Z] = \alpha Y + (1-\alpha) Z.\tag{1}$$
The right side of $(1)$ is the minimum-mean-square-error (MMSE) estimator of $X$ given $Y$ and $Z$. Since the estimator is a linear function of $Y$ and $Z$, 
it is also the linear MMSE estimator $L(X\mid Y,Z)$ for $X$ in terms of $Y$ and $Z$. Note that $E[\alpha Y + (1-\alpha) Z] = \mu = E[X]$ and thus $X-(\alpha Y + (1-\alpha) Z)$ is a zero-mean random variable. Hence,
$$\begin{align}
E[(X-(\alpha Y + (1-\alpha) Z))^2]
&=\operatorname{var}(X-(\alpha Y + (1-\alpha) Z))\\
&=\operatorname{var}(X-(\alpha (X+A) + (1-\alpha) (X+B))\\
&= \operatorname{var}(-\alpha A + (1 - \alpha) B)\\
&= \alpha^2\operatorname{var}(A) + (1-\alpha)^2\operatorname{var}(B)\\
&= (\operatorname{var}A)+\operatorname{var}(B))\alpha^2
-2\alpha\operatorname{var}(B) + \operatorname{var}(B)
\end{align}$$
which has minimum value when $$\displaystyle \alpha = \frac{\operatorname{var}(B)}{\operatorname{var}A)+\operatorname{var}(B)},\quad 1-\alpha = \frac{\operatorname{var}(A)}{\operatorname{var}(A)+\operatorname{var}(B)},
\tag{2}$$
that is,
$$E[X\mid Y, Z] =  L(X\mid Y,Z) = \frac{\operatorname{var}(B)\cdot Y +\operatorname{var}(A)\cdot Z}{\operatorname{var}A)+\operatorname{var}(B)}$$
which matches the answer $\dfrac{Y\sigma_z^2 + Z\sigma_y^2}{\sigma_y^2 + \sigma_z^2}$ suggested in a comment by @Henry.  (In the notation
used by the OP, $\sigma_y^2$ and $\sigma_z^2$ are the variances of $A$ 
and $B$ respectively, not of $Y$ and $Z$ as one might  expect.)

The OP does not state anywhere that $X$ is a normal random variable as I
have assumed above. Now, if $X$ is not a normal random variable, then 
even if we continue to assume that $X$, $A$, and $B$ are independent
random variables, it is not true that $X,Y,Z$ are jointly normal
and the right side of $(1)$ is not necessarily the conditional
mean of $X$ given $Y$ and $Z$. However it is still true that
the right side of $(3)$ is the linear MMSE estimator $L(X\mid Y,Z)$ of $X$
 given $Y$ and $Z$; this property is not contingent on joint
normality of $X$, $A$ and $B$ or even of marginal normality of
any of these random variables. The MMSE estimator 
$E[X\mid Y, Z]$ is the same 
as the linear MMSE estimator $L(X\mid Y,Z)$ for jointly normal
random variables, but the two estimators are usually different 
when joint normality does not exist or cannot be assumed to hold.
For example, suppose that $X$ is uniformly distributed on $(0,1)$
and that
$$f_{X,Y,Z}(x,y,z) = \frac{1}{2\pi}\exp\left(-\frac{(y-x)^2+(z-x)^2}{2}\right)
\mathbf 1_{x\in (0,1)},$$
that is, conditioned on $X=x$, $Y$ and $Z$ are (conditionally)
independent unit-variance normal random variables with mean $x$.
Then, conditioned on $(Y,Z) = (y,z)$, the conditional density of
$X$ is 
$$f_{X\mid Y,Z}(x\mid Y=y, Z=z) 
= \frac{f_{X,Y,Z}(x,y,z)}{f_{Y,Z}(y,z)}\mathbf 1_{x\in (0,1)}.$$
Consequently,
$$E[X\mid Y=y,Z=z] 
= \int_0^1 x \cdot f_{X\mid Y,Z}(x\mid Y=y, Z=z) \,\mathrm dx = g(y,z)$$ 
where it must be that
$g(y,z) \in [0,1]$ for all $y$ and $z \in \mathbb R$.
Thus, $g(Y,Z) = E[X\mid Y,Z]$ is a random variable that takes on values
in $[0,1]$, and so $E[X\mid Y,Z]$ cannot be the same random variable
as the random variable $L(X\mid Y,Z) = \alpha Y + (1-\alpha)Z$ which takes
on all real values. In this case, we have

$$E[X\mid Y,Z] = g(Y,Z) \neq L(X\mid Y,Z) = \frac{\operatorname{var}(B)\cdot Y +\operatorname{var}(A)\cdot Z}{\operatorname{var}A)+\operatorname{var}(B)} = \frac{Y+Z}{2}.$$

that is, independence of $X$, $A$, and $B$ is not sufficient to
allow us to assert that $E[X\mid Y,Z]$ is the same as $L(X\mid Y,Z)$.
The law of iterated expectation tells us that
$E[E[X\mid Y,Z]] = E[X]$ which has value $\frac 12$. Note also
that $E[Y] = E[E[Y\mid X]]$, and since $E[Y\mid X] = X$, we get
that $E[Y] = E[X] = \frac 12$, and similarly $E[Z]= \frac 12$.
Hence, 
$$E[L(X\mid Y, Z] = E\left[\frac{Y+Z}{2}\right] = \frac{E[Y]+E[Z]}{2} 
= \frac 12.$$
