Consider independent random variables $X_1$, $X_2$, and $X_3$ such that $X_1$ is a random variable having mean $1$ and variance $1$, $X_2$ is a random variable having mean $2$ and variance $4$, and $X_3$ is a random variable having mean $3$ and variance $9$.
(a) Give the value of the variance of $X_1+\frac{1}{2}X_2+\frac{1}{3}X_3$.
(b) Give the value of the correlation of $Y=X_1−X_2$ and $Z=X_2+X_3$.
(a) $$\begin{align*} Var\left(X_1+\frac{1}{2}X_2+\frac{1}{3}X_3\right) &= 1^2Var(X_1)+\frac{1}{2}^2Var(X_2)+\frac{1}{3}^2Var(X_3) \\\\ &= 1^2\cdot1+\frac{1}{4}\cdot4+\frac{1}{9}\cdot9 \\\\ &=1+1+1\\\\ &= 3 \end{align*}$$
(b)
I have the following:
$E(Y)=E(X_1-X_2)=E(X_1)-E(X_2)=1-2=-1$
$Var(Y)=Var(X_1-X_2)=1^2Var(X_1)+{-1}^2Var(X_2)=Var(X_1)+Var(X_2)=1+4=5$
$E(Z)=E(X_2+X_3)=E(X_2)+E(X_3)=2+3=5$
$Var(Z)=Var(X_2+X_3)=1^2Var(X_2)+1^2Var(X_3)=Var(X_2)+Var(X_3)=4+9=13$
I also have the formula
$$\begin{align*} \rho_{YZ} &= Corr(Y,Z) \\\\ &= \frac{Cov(Y,Z)}{\sigma_Y\sigma_Z} \\\\ &= \frac{Cov(Y,Z)}{\sqrt{5}\cdot\sqrt{13}} \\\\ \end{align*}$$
Is what I have so far correct? How could I use the information provided to find $Cov(Y,Z)$?
I tried
$$\begin{align*} Cov(Y,Z) &= E(YZ)-E(Y)E(Z) \\\\ &= E((X_1-X_2)(X_2+X_3))-E(X_1-X_2)\cdot E(X_2+X_3) \\\\ &= E(X_1\cdot X_2 - X_2^2 +X_1\cdot X_3 -X_2\cdot X_3)-E(X_1-X_2)\cdot E(X_2+X_3) \\\\ &= E(X_1\cdot X_2) - E(X_2^2) +E(X_1\cdot X_3) -E(X_2\cdot X_3))-E(X_1-X_2)\cdot E(X_2+X_3) \\\\ &=(1\cdot2) - (2^2) + (1\cdot 3) - (2\cdot3) - (-1\cdot 5) \\\\ &= 0 \end{align*}$$
Apply the Bilinearity of Covariance: $$\def\Cov{\mathsf{Cov}}\def\Var{\mathsf{Var}} {\Cov(Y,Z) ~{=~\Cov(X_1-X_2, X_2+X_3) \\=~ \Cov(X_1,X_2)+\Cov(X_1,X_3)-\Cov(X_2,X_2)-\Cov(X_2,X_3)\\=~0+0-\Var(X_2)-0\\ =~ -4}}$$