It can be factorised as $(x+y)^2+(y+z)^2+(z+x)^2=12$ $12$ can be written as some of $3$ squares as $4+4+4$ only The sums $x+y$, $y+z$ & $z+x$ can take values $2$ or $-2$ Solutions are $(x,y,z)=(-1,-1,3), (1,1,-3), (1,1,1), (-1,-1,-1)$ and its permutations

$x^2+y^2+z^2+xy+yz+zx=6$ $\Rightarrow 2(x^2+y^2+z^2+xy+yz+zx)=12$ $\Rightarrow (x+y)^2+(y+z)^2+(z+x)^2=12=3 \cdot 4$ $\Rightarrow (x+y)^2=(y+z)^2=(z+x)^2=4$ $\Rightarrow x+y= \pm 2 \wedge y+z= \pm 2 \wedge z+x= \pm 2$ $\Rightarrow (x,y,z)=(1,1,1),(-1,-1,-1),(-1,-1,3),(-1,3,-1),(3,-1,-1),(1,1,-3),(1,-3,1),(-3,1,1)$ $\blacksquare$

tarzanjunior wrote: It can be factorised as $(x+y)^2+(y+z)^2+(z+x)^2=12$ $12$ can be written as some of $3$ squares as $4+4+4$ only The sums $x+y$, $y+z$ & $z+x$ can take values $2$ or $-2$ Solutions are $(x,y,z)=(-1,-1,3), (1,1,-3), (1,1,1), (-1,-1,-1)$ and its permutations Great Solution!