In particular, $x+y+z$ is odd (a sum of 3 odd numbers). If $x+y+z\equiv 2\ (\text{mod}\ 3)$, there is an odd prime $p\equiv 2\ (\text{mod}\ 3)$ dividing $x+y+z$, and thus also \[0\equiv x^2+y^2+z^2\equiv x^2+y^2+(x+y)^2= 2(x^2+xy+y^2)\equiv x^2+xy+y^2\ (\text{mod}\ p)\]. Because $-3$ is not a quadratic residue mod $p$, this implies $x\equiv y\equiv 0(\equiv z)\ (\text{mod}\ p)$, contradicting $\text{gcd}(x,y,z)=1$.