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 \( x_{1}+x_{2}=4 \), the perpendicular bisector of segment \( AB \) intersects the \( x \)-axis at point \( C \). Find the maximum value of the area of \( \triangle ABC \).
10. As shown in Figure 3, let the midpoint of line segment $AB$ be $M\left(x_{0}, y_{0}\right)$.
Then
$$
\begin{array}{l}
x_{0}=\frac{x_{1}+x_{2}}{2}=2 \text {, } \\
y_{0}=\frac{y_{1}+y_{2}}{2} \text {, } \\
k_{A B}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\
=\frac{y_{2}-y_{1}}{\frac{y_{2}^{2}}{6}-\frac{y_{1}^{2}}{6}}=\frac{6}{y_{2}+y_{1}}=\frac{3}{y_{0}} \text {. } \\
\end{array}
$$

The equation of the perpendicular bisector of line segment $AB$ is
$$
y-y_{0}=-\frac{y_{0}}{3}(x-2) \text {. }
$$

It is easy to see that $C(5,0)$.
From equation (1), the equation of line $AB$ is
$$
\boldsymbol{y}-y_{0}=\frac{3}{y_{0}}(x-2) \text {, }
$$

which is $x=\frac{y_{0}}{3}\left(y-y_{0}\right)+2$.
Substituting equation (2) into $y^{2}=6 x$ gives
$$
y^{2}=2 y_{0}\left(y-y_{0}\right)+12 \text {, }
$$

which is $y^{2}-2 y_{0} y+2 y_{0}^{2}-12=0$.
According to the problem, $y_{1}$ and $y_{2}$ are the two distinct real roots of equation (3). Therefore,
$$
\Delta=4 y_{0}^{2}=4\left(2 y_{0}^{2}-12\right)=-4 y_{0}^{2}+48>0 \text {. }
$$

Solving this, we get $-2 \sqrt{3}<y_{0}<2 \sqrt{3}$.
$$
\begin{array}{l}
\text { Then }|A B|=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}} \\
=\sqrt{\left[1+\left(\frac{y_{0}}{3}\right)^{2}\right]\left(y_{1}-y_{2}\right)^{2}} \\
=\sqrt{\left(1+\frac{y_{0}^{2}}{9}\right)\left[\left(y_{1}+y_{2}\right)^{2}-4 y_{1} y_{2}\right]} \\
=\frac{2}{3} \sqrt{\left(9+y_{0}^{2}\right)\left(12-y_{0}^{2}\right)} .
\end{array}
$$

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

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

The equality holds if and only if $9+y_{0}^{2}=24-2 y_{0}^{2}$, i.e., $y_{0}= \pm \sqrt{5}$.
Therefore, the maximum area of $\triangle A B C$ is $\frac{14 \sqrt{7}}{3}$.