Saudi Arabian Mathematical Competitions

We can write
$$x^{2}+y^{2}+z^{2}+1=xy+yz+zx+|x-y+z-y|$$
hence
$$(x-y)^{2}+(y-z)^{2}+(z-x)^{2}+2=2|x-y+z-y| .$$
It follows
$$(x-y)^{2}+(y-z)^{2}+(z-x)^{2}+2 \leq 2|x-y|+2|y-z| .$$
The last relation is equivalent to
$$(|x-y|-1)^{2}+(|y-z|-1)^{2}+(z-x)^{2} \leq 0 .$$
We get $|x-y|=1$, $|y-z|=1$ and $x=z$. The desired triples $(x, y, z)$ are $(a, a-1, a)$, $(a, a+1, a)$, where $a \in \mathbb{R}$.