9. (16 points) For a given $m \in[-\sqrt{3}, \sqrt{3}]$, find the range of real numbers $n$ such that for any $x, y, z \in \mathbf{R}$, the following inequality always holds:
$$
x^{2}+2 y^{2}+3 z^{2}+2 x y+2 m z x+2 n y z \geqslant 0 \text {. }
$$
$$
\begin{array}{l}
x^{2}+2 y^{2}+3 z^{2}+2 x y+2 m z x+2 n y z \\
=(x+y+m z)^{2}+y^{2}+\left(3-m^{2}\right) z^{2}+(2 n-2 m) y z \\
\geqslant 0 .
\end{array}
$$
(1) When $3-m^{2}>0$,
$$
\begin{array}{l}
\Delta=4(n-m)^{2}-4\left(3-m^{2}\right) \leqslant 0 \\
\Rightarrow(n-m)^{2} \leqslant 3-m^{2} \\
\Rightarrow m-\sqrt{3-m^{2}} \leqslant n \leqslant m+\sqrt{3-m^{2}} .
\end{array}
$$
(2) When $3-m^{2}=0$,
$$
2 n-2 m=0 \Rightarrow n=m= \pm \sqrt{3} \text {. }
$$

In summary, $m-\sqrt{3-m^{2}} \leqslant n \leqslant m+\sqrt{3-m^{2}}$.