$\Longrightarrow 2x^2+2y^2+2z^2-2xy-2xz-2yz= (x-y)^2+(x-z)^2+(y-z)^2=6=1+1+4$ $\Longrightarrow (x,y,z)=(t,t+1,t+2), (t,t-1,t-2),~~~ t\in Z_0$ and permutations thereof.

$x^2+y^2+z^2-xy-yz-zx=(y-x)^2+(z-y)(z-x)=3$ and assume $x\leq y \leq z$. Then we realize that either $y-x=0$ and $z-y$ and $z-x$ must be $1$ and $3$ in some order which is impossible or $y-x=1$ and $z-y$ and $z-x$ must be $1$ and $2$ in some order which works but only if $z-y=y-x=1$, so the $3$ numbers must be $m-1$ $m$ and $m+1$ for some integer $m$. Plugging this in we realize all $m$ work and done.

re-express the equation as $(x+y+z)^2-3(xy+yz+zx)=3$. Clearly $x+y+z$ is a multiple of $3$ so $(x+y+z)^2$ is a multiple of $9$. This means that $xy+yz+zx$ has to be $2$ in modular $3$. Doing casework we realize that ${x mod 3, y mod 3, z mod 3}={0,1,2}$. Not sure where to go with this though.