10. 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 line segment $A B$ intersects the $x$-axis at point $C$. Find the maximum value of the area of $\triangle A B C$.
$$
k_{A B}=\frac{y_{1}-y_{2}}{x_{1}-x_{2}}=\frac{6}{y_{1}+y_{2}} \Rightarrow A B: y=\frac{6 x+y_{1} y_{2}}{y_{1}+y_{2}},
$$

Thus, the midpoint $D\left(2, \frac{12+y_{1} y_{2}}{y_{1}+y_{2}}\right)$ of segment $A B$.
Therefore, $D C: y-\frac{12+y_{1} y_{2}}{y_{1}+y_{2}}=-\frac{y_{1}+y_{2}}{6}(x-2)$,
Let $y=0 \Rightarrow x=\frac{12+y_{1} y_{2}}{\frac{y_{1}^{2}+y_{2}^{2}}{6}+\frac{y_{1} y_{2}}{3}}+2=5 \Rightarrow C(5,0)$.
Thus, $\overrightarrow{C A}=\left(x_{1}-5, y_{1}\right), \overrightarrow{C B}=\left(x_{2}-5, y_{2}\right)$,
where $y_{1}^{2}+y_{2}^{2}=6\left(x_{1}+x_{2}\right)=24$.
$S_{\triangle A B C}=\frac{1}{2}\left|\left(x_{1}-5\right) y_{2}-\left(x_{2}-5\right) y_{1}\right|=\frac{1}{2}\left|\frac{y_{1}^{2} y_{2}}{6}-\frac{y_{1} y_{2}^{2}}{6}+5\left(y_{1}-y_{2}\right)\right|$
$=\frac{1}{2}\left|y_{1}-y_{2}\right|\left|\frac{y_{1} y_{2}}{6}+5\right|=\frac{1}{12} \sqrt{y_{1}^{2}+y_{2}^{2}-2 y_{1} y_{2}}\left|y_{1} y_{2}+30\right|$
$=\frac{1}{12} \sqrt{\left(24-2 y_{1} y_{2}\right)\left(y_{1} y_{2}+30\right)^{2}} \leqslant \frac{1}{12} \sqrt{\left(\frac{84}{3}\right)^{3}}=\frac{1}{12} \cdot 28 \cdot 2 \sqrt{7}=\frac{14}{3} \sqrt{7}$.
Equality holds when $\left\{\begin{array}{l}24-2 y_{1} y_{2}=y_{1} y_{2}+30 \Rightarrow y_{1} y_{2}=-2, \\ y_{1}^{2}+y_{2}^{2}=24\end{array} \Rightarrow\left\{\begin{array}{l}y_{1}=\sqrt{5}-\sqrt{7}, \\ y_{2}=\sqrt{5}+\sqrt{7} .\end{array}\right.\right.$ Hence, the maximum area of $\triangle A B C$ is $\frac{14}{3} \sqrt{7}$.