10. (20 points) As shown in Figure 2, the ellipse $C: \frac{x^{2}}{4}+\frac{y^{2}}{3}=1$ has left and right vertices $A$ and $B$, respectively. Take any point $P$ on the ellipse $C$ different from $A$ and $B$. The lines $PA$ and $PB$ intersect the line $x=3$ at points $M$ and $N$, respectively. The line $MB$ intersects the ellipse $C$ at point $Q$.
(1) Find the value of $\overrightarrow{F M} \cdot \overrightarrow{F N}$;
(2) Prove that points $A$, $Q$, and $N$ are collinear.
10. (1) Let point $P\left(x_{0}, y_{0}\right)$. Then $3 x_{0}^{2}+4 y_{0}^{2}=12$. From $l_{P A}: y=(x+2) \frac{y_{0}}{x_{0}+2}$, we get $M\left(3, \frac{5 y_{0}}{x_{0}+2}\right)$;
From $l_{P B}: y=(x-2) \frac{y_{0}}{x_{0}-2}$, we get $N\left(3, \frac{y_{0}}{x_{0}-2}\right)$.
Then, from point $F(1,0)$, we have
$$
\begin{array}{l}
\overrightarrow{F M} \cdot \overrightarrow{F N}=\left(2, \frac{5 y_{0}}{x_{0}+2}\right) \cdot\left(2, \frac{y_{0}}{x_{0}-2}\right) \\
=4+\frac{5 y_{0}^{2}}{x_{0}^{2}-4}=4-\frac{15}{4}=\frac{1}{4} .
\end{array}
$$
(2) Let point $Q$ be $P^{\prime}(s, t)$.

From (1), we know that the line $P^{\prime} B$ intersects the line $x=3$ at point $N^{\prime}\left(3, \frac{t}{s-2}\right)$, and the line $P^{\prime} A$ intersects the line $x=3$ at point $M^{\prime}\left(3, \frac{5 t}{s+2}\right)$.
By the problem, point $N^{\prime}$ is point $M$, so
$$
\frac{t}{s-2}=\frac{5 y_{0}}{x_{0}+2} \text {. }
$$

Furthermore, from $\frac{t}{s-2} \cdot \frac{5 t}{s+2}=-\frac{15}{4}=\frac{5 y_{0}}{x_{0}+2} \cdot \frac{y_{0}}{x_{0}-2}$, we get
$$
\frac{5 t}{s+2}=\frac{y_{0}}{x_{0}-2} \text {. }
$$

Thus, point $M^{\prime}$ is point $N$.
Therefore, from the collinearity of points $A, P^{\prime}, M^{\prime}$, we get the collinearity of points $A, Q, N$.