10. (20 points) Given two moving points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ on the parabola $y^{2}=6 x$, where $x_{1} \neq x_{2}$ and $x_{1}+x_{2}=4$. If the perpendicular bisector of segment $A B$ intersects the $x$-axis at point $C$, find the maximum value of the area of $\triangle A B C$.
10. Let the midpoint of line segment $AB$ be $M\left(x_{0}, y_{0}\right)$. Then $x_{0}=\frac{x_{1}+x_{2}}{2}=2, y_{0}=\frac{y_{1}+y_{2}}{2}$.

Thus, $k_{AB}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{y_{2}-y_{1}}{\frac{y_{2}^{2}}{6}-\frac{y_{1}^{2}}{6}}=\frac{6}{y_{2}+y_{1}}=\frac{3}{y_{0}}$.
The equation of the perpendicular bisector of line segment $AB$ is
$$
y-y_{0}=-\frac{y_{0}}{3}(x-2),
$$

and $x=5, y=0$ is a solution to equation (1), so point $C(5,0)$.
From equation (1), we have
$$
\begin{array}{l}
l_{AB}: y-y_{0}=\frac{3}{y_{0}}(x-2) \\
\Rightarrow x=\frac{y_{0}}{3}\left(y-y_{0}\right)+2,
\end{array}
$$

Substituting into $y^{2}=6 x$ gives
$$
\begin{array}{l}
y^{2}=2 y_{0}\left(y-y_{0}\right)+12 \\
\Rightarrow y^{2}-2 y_{0} y+2 y_{0}^{2}-12=0 .
\end{array}
$$

According to the problem, $y_{1}$ and $y_{2}$ are the two distinct real roots of equation (2).
Then $\Delta=4 y_{0}^{2}-4\left(2 y_{0}^{2}-12\right)=-4 y_{0}^{2}+48>0$
$\Rightarrow-2 \sqrt{3}<y_{0}<2 \sqrt{3}$.
Thus, $|AB|=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
$$
\begin{array}{l}
=\sqrt{\left(1+\left(\frac{y_{0}}{3}\right)^{2}\right)\left(y_{1}-y_{2}\right)^{2}} \\
=\sqrt{\left(1+\frac{y_{0}^{2}}{9}\right)\left(\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}\right)} \\
=\sqrt{\left(1+\frac{y_{0}^{2}}{9}\right)\left(4 y_{0}^{2}-4\left(2 y_{0}^{2}-12\right)\right)} \\
=\frac{2}{3} \sqrt{\left(9+y_{0}^{2}\right)\left(12-y_{0}^{2}\right)} .
\end{array}
$$

The distance from the fixed point $C(5,0)$ to line segment $AB$ is
$$
\begin{aligned}
h & =|CM|=\sqrt{(5-2)^{2}+\left(0-y_{0}\right)^{2}} \\
& =\sqrt{9+y_{0}^{2}},
\end{aligned}
$$

Thus, $S_{\triangle ABC}=\frac{1}{2}|AB| h$
$$
\begin{array}{l}
=\frac{1}{3} \sqrt{\left(9+y_{0}^{2}\right)\left(12-y_{0}^{2}\right)} \sqrt{9+y_{0}^{2}} \\
=\frac{1}{3} \sqrt{\frac{1}{2}\left(9+y_{0}^{2}\right)\left(24-2 y_{0}^{2}\right)\left(9+y_{0}^{2}\right)} \\
\leqslant \frac{1}{3} \sqrt{\frac{1}{2}\left(\frac{9+y_{0}^{2}+24-2 y_{0}^{2}+9+y_{0}^{2}}{3}\right)^{3}} \\
=\frac{14 \sqrt{7}}{3} .
\end{array}
$$

Equality in equation (3) holds if and only if $9+y_{0}^{2}=24-2 y_{0}^{2}$, i.e., $y_{0}= \pm \sqrt{5}$, $A\left(\frac{6+\sqrt{35}}{3}, \sqrt{5}+\sqrt{7}\right), B\left(\frac{6-\sqrt{35}}{3}, \sqrt{5}-\sqrt{7}\right)$ or $A\left(\frac{6+\sqrt{35}}{3},-(\sqrt{5}+\sqrt{7})\right), B\left(\frac{6-\sqrt{35}}{3},-\sqrt{5}+\sqrt{7}\right)$.
Therefore, the maximum area of $\triangle ABC$ is $\frac{14 \sqrt{7}}{3}$.