As it was proved in the answer here, for $Z_1, Z_2$, two independent and identically distributed random variables. Then we have: $$ \mathbb E[Z_1\mid Z_1+Z_2] =\frac{Z_1+Z_2}{2}. $$
However, suppose now that $Z_1, Z_2$ are two standard normal random variables and that we define $W_1 = aZ_1 \sim \mathcal N(0, a^2)$ and $W_2 = bZ_2 \sim \mathcal N(0, b^2)$.
Thus, $W_1, W_2$ are two still independent normal random variables, but with different distributions as stated. Question: what is $$ \mathbb E[Z_1\mid W_1+W_2]? $$
The problem is that now the argument in the question linked seems not to be working here, because now $W_1+W_2$ is not anymore a symmetric function of $(Z_1, Z_2)$.