7. Given positive integers $x, y, z$ satisfying $x y z=(14-x)(14-y)(14-z)$, and $x+y+z<28$. Then the maximum value of $x^{2}+y^{2}+z^{2}$ is . $\qquad$
7. 219 .

From the problem, we know that $x$, $y$, and $z$ are all positive integers less than 14. On the other hand, expanding the given equation, we get
$$
2 x y z=14^{3}-14^{2}(x+y+z)+14(x y+y z+z x) \text {. }
$$

Thus, $71 x y z$.
Since $x$, $y$, and $z$ are all less than 14, at least one of $x$, $y$, or $z$ must be 7.
Without loss of generality, let $z=7$. Then the given equation simplifies to
$$
x y=(14-x)(14-y) \Rightarrow x+y=14 \text {. }
$$

In this case, the maximum value of $x^{2}+y^{2}$ is $1^{2}+13^{2}$.
Therefore, the maximum value of $x^{2}+y^{2}+z^{2}$ is
$$
1^{2}+13^{2}+7^{2}=219 \text {. }
$$