Let $X_1, X_2, X_3$ be three standard normal variables, let let U be uniform on $\{ 1,2,3,4 \}$, and let $S_1 = 1 $ if $U \in \{ 1,2 \}$ and $-1$ if $U \in \{ 3,4 \}$, analogously $S_2 = 1$ if $U \in \{ 1,3 \}$ and $S_3 = 1$ if $U \in \{1,4 \}$. Let $Y_1 = S_1 | X_1 |$, and define analogously $Y_2, Y_3$.
Then we can prove that each $Y_i$ is Gaussian, and they are pairwise independent. But I'm supposed to find a linear combination of them that is not Gaussian.
Intuitively, even if we take $Y_1 + Y_2 + Y_3$, then for $U = 1,2,3$ we have a term like $|X_1| + |X_2| - |X_3|$, and for $U = 4$ we have $- |X_1| - |X_2| - |X_3|$, so there are three cases in which the sum "is positive by 1", and one case when it's "negative by 3" - so the expected value averages out to 0, as it should be if it was Gaussian. It seems that we should maybe look at the variance, because the squared distances are not uniform in this way any more - so the distribution would be skewed. But I don't know how I would actually finish that argument. Is that the right approach? Or is there some combination that gives a better solution?
It seems like this is similar to the problem of sum of two Gaussians where their sum is not Gaussian, but there you pick $X$ and $\pm X$ based on a coin flip - but here, we start with 3 independent Gaussians, so I don't see how that can be applied.
This is probably not what you are looking for, but there is a standard result that says that $Z_1,...,Z_n$ are jointly Gaussian iff any linear combination of the $Z_1,...,Z_n$ is Gaussian.
Hence if we can show that $Y_1,Y_2,Y_3$ are not jointly Gaussian, then it follows that there is some linear combination that is not Gaussian.
Suppose $Y=(Y_1,Y_2,Y_3)$ is jointly Gaussian. Since $E [Y_i Y_j] = \delta_{ij} $, we see that the covariance is positive definite, hence, for any open $U$, we have $P[Y \in U] >0$. However, if we let $U=B((1,-1,1), {1 \over 2})$, we see that $P[Y \in U] = 0$ which is a contradiction. Hence the $Y_1,Y_2,Y_3$ are not jointly Gaussian.
(Empirically, if you look at a bin plot of $Y_1+Y_2+Y_3$, it is obvious that it is not symmetric, but I wasn't able to turn this observation into a proof.)