11. Given positive integers $x, y, z$ satisfying $x y z=(22-x)(22-y)(22-z)$, and $x+y+z<44, x^{2}+y^{2}+z^{2}$, the maximum and minimum values are denoted as $M$ and $N$, respectively, then $M+N=$ $\qquad$.
Rong 926.
Analysis From the given information, we know that $x, y, z$ are all positive integers less than 22. From the given equation, we have
$$
2 x y z=22^{3}-22^{2}(x+y+z)+22(x y+y z+z x)
$$

Thus, $11 \mid x y z$, so at least one of $x, y, z$ must be equal to 11.
Assume without loss of generality that $x=11$. From the given equation, we get $y+z=22$. Since $y, z$ are both positive integers, we also have
$$
y^{2}+z^{2}=\frac{(y+z)^{2}+(y-z)^{2}}{2}=\frac{22^{2}+(y-z)^{2}}{2}=242+\frac{1}{2}(y-z)^{2}
$$

It is easy to see that when $y=z=11$, $y^{2}+z^{2}$ takes its minimum value of 242, and when $|y-z|=|21-1|=20$, $y^{2}+z^{2}$ takes its maximum value. In this case, $M=11^{2}+21^{2}+1^{2}=563$, so $M+N=563+363=926$.