12. Given the ellipse $\frac{x^{2}}{6}+\frac{y^{2}}{2}=1$ with its right focus at $F$, the line $y=k(x-2)$ through $F$ intersects the ellipse at points $P$ and $Q$ $(k \neq 0)$. If the midpoint of $P Q$ is $N$, and $O$ is the origin, the line $O N$ intersects the line $x=3$ at $M$.
(1) Find the size of $\angle M F Q$;
(2) Find the maximum value of $\frac{|P Q|}{|M F|}$.
(1) As shown in the figure, solving the system $\left\{\begin{array}{l}y=k(x-2), \\ x^{2}+3 y^{2}=6\end{array}\right.$
$$
\Rightarrow\left(3 k^{2}+1\right) x^{2}-12 k^{2} x+12 k^{2}-6=0,
$$

we get $N\left(\frac{6 k^{2}}{3 k^{2}+1},-\frac{2 k}{3 k^{2}+1}\right)$.
Thus, $k_{O N}=-\frac{1}{3 k} \Rightarrow$ the line $O N: y=-\frac{x}{3 k}$, hence $M\left(3,-\frac{1}{k}\right) \Rightarrow k_{M F}=-\frac{1}{k}$.
Therefore, $M F \perp P Q \Rightarrow \angle M F Q=90^{\circ}$.
(2) Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$, then $|P Q|=\sqrt{k^{2}+1} \cdot \frac{\sqrt{12^{2} k^{4}-24\left(3 k^{2}+1\right)\left(2 k^{2}-1\right)}}{3 k^{2}+1}$ $=\frac{2 \sqrt{6}\left(k^{2}+1\right)}{3 k^{2}+1},|M F|=\sqrt{1+\frac{1}{k^{2}}}=\frac{\sqrt{k^{2}+1}}{|k|}$, thus $\frac{|P Q|}{|M F|}=\frac{2 \sqrt{6}|k| \cdot \sqrt{k^{2}+1}}{3 k^{2}+1}$ $=\frac{2 \sqrt{3} \cdot \sqrt{2}|k| \cdot \sqrt{k^{2}+1}}{3 k^{2}+1} \leqslant \frac{2 \sqrt{3} \cdot \frac{2 k^{2}+k^{2}+1}{2}}{3 k^{2}+1}=\sqrt{3}$, equality holds when $k^{2}=1$.
Therefore, the maximum value of $\frac{|P Q|}{|M F|}$ is $\sqrt{3}$.