(7) Given positive integers $x, y, z$ satisfy the condition $x y z=(14-x)(14-y)(14-$
$z)$, and $x+y+z<28$, then the maximum value of $x^{2}+y^{2}+z^{2}$ is $\qquad$.
(7) 219 Hint: Given that $x, y, z$ are positive integers,
$$
(14-x)(14-y)(14-z)=x y z>0,
$$

and $x+y+z<28$, we know that $x, y, z$ are all positive integers less than 14.
On the other hand, expanding
$$
x y z=(14-x)(14-y)(14-z)
$$

we get
$$
x y z=14^{3}-14^{2}(x+y+z)+14(x y+y z+z x)-x y z,
$$

which simplifies to
$$
2 x y z=14^{3}-14^{2}(x+y+z)+14(x y+y z+z x) .
$$

Therefore, $x y z$ is divisible by 7. Given that 7 is a prime number and $x, y, z$ are positive integers less than 14, at least one of $x, y, z$ must be 7. Without loss of generality, let $z=7$, then the condition equation becomes
$$
x y=(14-x)(14-y) .
$$

Thus,
$$
x y=14^{2}-14(x+y)+x y,
$$

which implies $x+y=14$. 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 $1^{2}+13^{2}+7^{2}=219$.