Let $X$, $Y$, and $Z$ be Gaussian random variables that are not independent. This does not imply that $(X,Y,Z)$ is a Gaussian vector. But if we add the conditions that $E[X|Y,Z]$ is a linear function of $Y,Z$, i.e, $E[X|Y,Z]=a+bY+cZ$ for some real numbers $a,b,c$, does that mean that $(X,Y,Z)$ is a Gaussian vector ?
The converse is true: If $(X,Y,Z)$ is Gaussian, then $E[X|Y,Z]$ is a linear function of $Y,Z$.
Here is an example of a triple of random variables $(X,Y,Z)$ such that each of the components $X$, $Y$, and $Z$ are individually gaussian but the three of them are not jointly gaussian. Further, the conditional expectation $E[X|Y,Z]=Z$, a linear function of $Y,Z$.
Let $U$ be a standard gaussian, let $V$ be independent of $U$ such that $P(V=1)=P(V=-1)=1/2$, let $Z=U$, let $Y=UV$, let $X=Z$. Then $E[X|Y,Z] = Z$, which is a linear function of the pair $Y,Z$. Clearly $X$ and $Z$ are standard gaussian. Since the distribution of $U$ is symmetric about $0$, the distribution of $U$ is the same as that of $-U$, and so is the distribution of $Y=UV$.