Let $X_1, X_2, X_3$ be iid $N(\mu,\sigma^2)$ random variables. Find $E[2X_1+3X_2|X_1+3X_2-X_3=4]$.
Inuitively, I think this expectation will be equal to $E[4+X_1+X_3]$, hence $4+2\mu$. As $2X_1+3X_2 = 4 + X_1+ X_3$.
Is my approach correct? Also, how do I show this mathematically?
On
Right for the wrong reason. Let $Y = X_1 + 3 X_2 - X_3$. Note that with the same reasoning you could have added any other multiple of $Y - 4$ to $2 X_1 + 3 X_2$ to get a different answer. The difference is that the covariance of $X_1 + X_3$ and $Y$ is zero, making them independent as uncorrelated components of a multinormal distribution. Therefore $$\operatorname E[2 X_1 + 3 X_2 \mid Y = 4] = \operatorname E[X_1 + X_3 + 4 \mid Y = 4] = \operatorname E[X_1 + X_3 + 4] = 2 \mu + 4.$$
The right way to solve this problem is to find the parameters of the joint bivariate normal distribution of $(U,V)=(2X_1+3X_2, X_1+3X_2-X_3)$. Then you can use a standard formula for conditional distributions for bivariate normal distributions. Or, if you know about linear regression, you could find the least squares fit of form $U=a + bV +\text{ error}$ and obtain your answer $E(U|V=4)=a+4b$.
There might be shorter or more intuitive ways to derive the answers to such problems but they are typically error prone and hard to explain to others. The present method is methodical and easy to check and easy to explain.