HMMT-Nov 2021 guts, Problem 25

Let $x, y, z$ be real numbers satisfying \[ \begin{aligned} 2 x+y+4 x y+6 x z & =-6 \\ y+2 z+2 x y+6 y z & =4 \\ x-z+2 x z-4 y z & =-3 \end{aligned} \] Find $x^{2}+y^{2}+z^{2}$.

Solution: Solution: We multiply the first, second, and third equations by $\frac{1}{2},-\frac{1}{2}$, and -1 , respectively, then add the three resulting equations. This gives $x y+x z+y z=-2$. Doing the same with the coefficients $-1,2$, and 3 gives $x+y+z=5$, from which $(x+y+z)^{2}=25$. So $x^{2}+y^{2}+z^{2}=25-2 \cdot-2=29$. $\fbox{29}$.

Answer: 29