Joint distribution of normally distributed random variables Let $X_1, X_2, X_3 \sim N(0,1)$ be independent random variables. I needed to figure out the distribution of $U=3X_1-X_2-2X_3+1$, $V=X_1+X_2+X_3$ and $W=2X_1-3X_2+X_3$ which resulted in $U \sim N(1,1)$, $V \sim N(0,3), W \sim N(0,2)$. Is that correct?
Now I need to calculate the joint distribution of $(U,V,W)$ which I think is done by drawing up the covariance matrix.
But I am not sure how to calculate $E(UV)$ for example which I need vor $cov(U,V)$. Any hints or approaches are much appreciated.
 A: Weighted Sum of Normal Distributions
Let $$Y = \alpha_1 X_1 + \ldots +\alpha_n X_n + \beta$$
where $X_k \sim N(0,1)$ then $Y$ is also normally distributed with mean
$$E(Y) = E(\alpha_1 X_1 + \ldots +\alpha_n X_n) = \alpha_1 E(X_1) + \ldots +\alpha_n E(X_n) = 0 + \ldots + 0+ \beta = \beta$$
and
$$\operatorname{var}(Y) = \operatorname{var}(\alpha_1 X_1 + \ldots +\alpha_n X_n+ \beta) = \alpha_1^2 \operatorname{var}(X_1) + \ldots +\alpha_n^2 \operatorname{var}(X_n)+0 = \alpha_1^2 + \ldots + \alpha_n^2$$
So your means are correct, but your variances are not. For example, 
$$\operatorname{var}W=2^2+3^2 +1^2=14$$
and the same applies for the rest.

Joint Distribution
The joint distribution of Normal distributions is a multi-variate normal with mean vector containing the means of each individual normal random variable, therefore you don't need $E(UV)$, you need $E(U), E(V),E(W)$, which, if placed in a vector, is $$\mu = [1, 0,0]$$
As for the covariance, you have $\operatorname{var}(U),\operatorname{var}(V)$ and $\operatorname{var}(W)$, which serve as the diagonal elements of the covariance matrix, now the off-diagonal element, which are $cov(U,V),cov(U,W),cov(V,W)$, you would have to compute them by definition, here's one example
\begin{equation}
\begin{split}
cov(U,V) &= E((U - E(U))(V-E(V)) \\
&= E(UV)\\
&= E(3X_1-X_2-2X_3+1)(X_1+X_2+X_3)\\
&= E(3X_1^2 + 3X_1X_2 + 3X_1X_3) \\
&- E(X_1X_2 + X_2^2 +X_1X_3) \\
&- E(2X_1X_3 + 2X_2X_3 + 2X_3^2)\\
&+ E(X_1+X_2+X_3)
\end{split}
\end{equation}
Notice that all terms $E(X_iX_j) =E(X_i)E(X_j) = 0$ due to independence, so
\begin{equation}
\begin{split}
cov(U,V) 
&= E(3X_1^2+X_2^2+2X_3^2+X_1+X_2+X_3) \\
&= 3E(X_1^2)  - E(X_2^2) - 2E(X_3^2) + E(X_1) + E(X_2) + E(X_3)\\
&= 3 - 1 - 2 + 0 + 0 + 0\\
&= 0
\end{split}
\end{equation}
A: You have $\boldsymbol X =(X_1, X_2, X_3)^t$ jointly normal with the mean vector $\boldsymbol \mu = 0$ and the covariance matrix $M = (\delta_{ij})$. $A \boldsymbol X + \boldsymbol b$ will also be jointly normal with $\boldsymbol \mu' = A \boldsymbol \mu + \boldsymbol b, \,M' = A M A^t$:
$$\begin{pmatrix} 3 & -1 & -2 \\ 1 & 1 & 1 \\ 2 & -3 & 1 \end{pmatrix}
 \boldsymbol X + \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \sim
\mathcal N \!\left( \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},
 \begin{pmatrix} 14 & 0 & 7 \\ 0 & 3 & 0 \\ 7 & 0 & 14 \end{pmatrix} \right).$$
