Let $x=12 x_{1}+r_{1}\left(0 \leqslant r_{1} \leqslant 11\right)$,
$$\begin{array}{l} y=4 y_{1}+r_{2}\left(0 \leqslant r_{2} \leqslant 3\right), \\ z=3 z_{1}+r_{3}\left(0 \leqslant r_{3} \leqslant 2\right) . \end{array}$$

The equation becomes
$$\begin{array}{ll} & 12\left(x_{1}+y_{1}+z_{1}\right)+r_{1}+3 r_{2}+4 r_{3}=665 . \\ & \therefore r_{1}+3 r_{2}+4 r_{3} \equiv 665 \equiv 5(\bmod 12) . \\ & \because 0 \leqslant r_{1}+3 r_{2}+4 r_{3} \leqslant 28, \\ & r_{1}+3 r_{2}+4 r_{3}=5 \\ \text { or } & r_{1}+3 r_{2}+4 r_{3}=17 . \end{array}$$

It is easy to find that (1) has 3 solutions, (2) has 8 solutions.
$\therefore$ The original equation has $3 C_{\frac{655 ~ 5}{2}-2}^{12}+8 C_{\frac{665-5}{2}-1+2}^{12}$ $=17108$ non-negative integer solutions.