$ Cov(X,Y)=-5, \, Cov(X,Z)=2, \, Cov(Y,Z)=2 $. What can we say about $\sigma^2 (X+Y+Z)$?

Question

I found this question in some exam, but i don't have the answer. I tried to solve it myself, but i'm afraid i got stuck.

We these random $3$ variables $X,Y,Z$, and we know that: $$ Cov(X,Y)=-5, \quad Cov(X,Z)=2, \quad Cov(Y,Z)=2 $$ Select the correct answer:

1. $\sigma^2 (X+Y+Z) < \sigma^2(X) + \sigma^2(Y) - \sigma^2(Z)$
2. $\sigma^2 (X+Y+Z) = \sigma^2(X) + \sigma^2(Y) + \sigma^2(Z)$
3. $\sigma^2 (X+Y+Z) > \sigma^2(X) + \sigma^2(Y) + \sigma^2(Z)$
4. $\sigma^2 (X-Y+Z) < \sigma^2(X) + \sigma^2(Y) + \sigma^2(Z)$
5. $\sigma^2 (-X+Y+Z) > \sigma^2(X) + \sigma^2(Y) + \sigma^2(Z)$

$$$$ Here's what i calculated so far:

First of all, i know that: $\sigma^2 (\pm X \pm Y \pm Z)=\sigma^2 (X) + \sigma^2(Y) + \sigma^2( Z)$, so i don't mind about the signs so that's why option $1$ is out (yeah, particularly now i just ignored the fact they are correlated. Of course in our case i must include to covariances).

Now, i tried to "split" the variance of $\sigma^2 (X+Y+Z)$ in $3$ different way (i skipped some steps, represented as "..."):

1. $\sigma^2 ((X+Y)+Z)=\sigma^2(X+Y)+\sigma^2(Z)+2Cov(X+Y,Z)=...= \sigma^2(X) + \sigma^2(Y) + \sigma^2(Z)+\boldsymbol{(2Cov(X+Y,Z)-10)}$
2. $\sigma^2 (X+(Y+Z))=\sigma^2(X)+\sigma^2(Y+Z)+2Cov(X,Y+Z)=...= \sigma^2(X) + \sigma^2(Y) + \sigma^2(Z)+\boldsymbol{(4+2Cov(X,Y+Z))}$
3. $\sigma^2 ((X+Z)+Y)=\sigma^2(X+Z)+\sigma^2(Y)+2Cov(X+Z,Y)=...= \sigma^2(X) + \sigma^2(Y) + \sigma^2(Z)+\boldsymbol{(4+2Cov(X+Z,Y)}$

and since all these $3$ bold parts are actually equal, i managed to conclude that: $$ Cov(X,Y+Z)= Cov(X+Z,Y), \quad Cov(X+Y,Z)=7+Cov(X+Z,Y)=7+Cov(X,Y+Z) $$

But i don't really see how can i move forward from here. What do you think the correct answer is? Can you give me a lead or a hint? Do you agree with my conclutions so far?