If the random variables $X,Y$ and $Z$ have means: $$\bar{x}=2 \;\;\; \bar{y}=-3 \;\;\; \bar{z}=4$$ and variances $$\operatorname{Var}(x)=1 \;\;\; \operatorname{Var}(Y)=5 \;\;\; \operatorname{Var}(Z)=2$$ and co-variance $$\operatorname{Cov}(X,Y)= -2\;\;\; \operatorname{Cov}(X,Z)=-1\;\;\;\operatorname{Cov}(Y,Z)=1$$ Find
$$E[aX+b] = aE[X]+b$$ $$\operatorname{Var}[aX+bY] = \mathbf{v}^T\mathbf{M}\mathbf{v}$$ where $\mathbf{v} = \left(\begin{matrix}a\\b\end{matrix}\right)$ and $\mathbf{M}$ is the matrix of covariance defined (in the case two variables) as $$\left(\begin{matrix}\operatorname{Var}[X]&\operatorname{Cov}[X,Y]\\\operatorname{Cov}[Y,X]&\operatorname{Var}[Y]\end{matrix}\right)$$
The fist question is the easiest, from the linearity of the mean you can simply add together the means with their respective coefficientent given by the formula for $W$, mainly $$E[W] = E[3X-Y+2Z] = 3E[X]-E[Y]+2E[Z] = 3(2)-(-3)+2(4) = 17$$ The second, given the formula above is to be solved by simple matrix multiplication. In our case $$\mathbf{v} = \left(\begin{matrix}3\\-1\\2\end{matrix}\right)\;\;\;\mathbf{M}=\left(\begin{matrix}\operatorname{Var[X]}&\operatorname{Cov[X,Y]}&\operatorname{Cov[X,Z]}\\\operatorname{Cov[Y,X]}&\operatorname{Var[Y]}&\operatorname{Cov[Y,Z]}\\\operatorname{Cov[Z,X]}&\operatorname{Cov[Z,Y]}&\operatorname{Var[Z]}\end{matrix}\right) = \left(\begin{matrix}1&-2&-1\\-2&5&1\\-1&1&2\end{matrix}\right)$$ Then we calculate the variance of $W$ as $$\operatorname{Var}[W] = \operatorname{Var}[3X-Y+2Z] = \underbrace{\left(\begin{matrix}3&-1&2\end{matrix}\right)\left(\begin{matrix}1&-2&-1\\-2&5&1\\-1&1&2\end{matrix}\right)}_{\text{first multiplication}}\left(\begin{matrix}3\\-1\\2\end{matrix}\right)$$ let's do first the underbraced multiplication $$\left(\begin{matrix}3&-1&2\end{matrix}\right)\left(\begin{matrix}1&-2&-1\\-2&5&1\\-1&1&2\end{matrix}\right) = \left(\begin{matrix}3&-9&6\end{matrix}\right)$$ and now let's do the second multiplication $$\left(\begin{matrix}3&-9&6\end{matrix}\right)\left(\begin{matrix}3\\-1\\2\end{matrix}\right)=30$$ So in the end we have that $$\operatorname{Var}[W]=30$$