6) Let the number of distinct positive integer solutions $(x, y, z)$ to the indeterminate equation $x^{2}+y^{2}+z^{2}-x y z+10=0$ where $x, y, z$ are all greater than 2008 be $k$, then $k$ satisfies ( ).
(A) $k=0$
(B) $1 \leqslant k \leqslant 2008$
(C) $k>2008$, but $k$ is a finite number
(D) $k$ is infinite
6 From the problem, we know that $\left(x_{0}, y_{0}, z_{0}\right)=(3,4,5)$ is a particular solution to the original indeterminate equation. For any positive integer solution $\left(x_{1}, y, z\right)$ of the original indeterminate equation, assuming $x_{1} \leqslant y \leqslant z$, and $x_{1}z^{2}$,

Therefore, $x_{2}$ is a positive integer, and greater than $z$, so $\left(x_{2}, y, z\right)$ is also a solution to the original indeterminate equation. Since the original indeterminate equation is cyclically symmetric, $\left(y, z, x_{2}\right)$ is also its solution, and it is arranged in ascending order. By repeatedly using the above conclusion, we can obtain infinitely many solutions from a particular solution, so there are infinitely many solutions satisfying $x, y, z$ all greater than 2008. Therefore, the answer is D.