11. Find all positive integers $x, y, z$, such that: $x y \pmod{z} \equiv y z \pmod{x} \equiv z x \pmod{y} = 2$.
11. Solution: From the conditions, we know that $x|y z-2, z| x y-2, y \mid x z-2$,

then $x y z \mid(y z-2)(x y-2)(x z-2)$,
and $(x y-2)(x z-2)(y z-2)=x^{2} y^{2} z^{2}-2\left(x^{2} y z+x y^{2} z+x y z^{2}\right)+4(x y+x z+y z)-8$, so $x y z \mid 4(x y+x z+y z)-8$.
Let $4(x y+x z+y z)-8=k x y z$, then $4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=k+\frac{8}{x y z}$.
From the conditions, we know that $x, y, z>2$, if two of $x, y, z$ are equal. Without loss of generality, let $x=y$, then $y z \equiv 0(\bmod x)$, which is a contradiction! Therefore, $x, y, z$ are distinct.
Without loss of generality, let $x \leqslant y \leqslant z$, then $4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \leqslant 4\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\right)\frac{4}{x y z}$, thus $\frac{4}{x}+\frac{4}{y}+\frac{4}{z}>1+\frac{8}{x y z}$, which is a contradiction!
If $x \geqslant 12$, then
$4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) \leqslant 4\left(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\right)<1$, which is a contradiction!
Therefore, $5 \leqslant x \leqslant 11$.
(¡) $x=5$, solving yields $(x, y, z)=(5,21,412),(5,22,216),(5,24,118),(5,28,69),(5,27,76),(5,34,48)$, none of which satisfy the requirements upon inspection.
(ii) $x=6$, substituting yields $(x, y, z)=(6,13,152),(6,14,82),(6,16,47),(6,17,40),(6,19,32),(6,22,26)$, upon inspection $(6,14,82),(6,22,26)$ satisfy the requirements.
(iii) $x=7$, solving yields $(x, y, z)=(7,10,136),(7,11,60),(7,12,41),(7,16,22)$, none of which satisfy the requirements upon inspection.
(iv) $x=8$, solving yields $(x, y, z)=(8,9,70),(8,10,39)$, none of which satisfy the requirements upon inspection.
Repeating the above process, it is easy to see that when $x=9,10,11$ none satisfy the requirements.
In summary, the $(x, y, z)$ that satisfy the conditions are $(3,8,22),(3,10,14),(4,5,18),(4,6,11),(6,14,82),(6,22,26)$ and their symmetric arrays.