6. Let the number of distinct positive integer solutions $(x, y, z)$ to the indeterminate equation $x^{2}+y^{2}+z^{2}-x y z+10=0$ that satisfy $x, y, z > 2008$ 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.D.
$\left(x_{0}, y_{0}, z_{0}\right)=(3,4,5)$ 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 $x_{1}z^{2} \text {. }$
 
Therefore, $x_{2}$ is a positive integer, and greater than $z$.
Thus, $\left(x_{2}, y, z\right)$ is also a solution to the original indeterminate equation, and when rearranged in ascending order, it becomes $\left(y, z, x_{2}\right)$.

By repeatedly applying the above conclusion, we can derive infinitely many solutions from a particular solution. Therefore, there are infinitely many solutions where $x, y, z$ are all greater than 2008.