Question 174, 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 $\mathrm{AB}$ intersects the $x$-axis at point $C$. Find the maximum value of the area of $\triangle \mathrm{ABC}$.
Given $x_{1}=2+t, x_{2}=2-t(t \in[-2,2])$, then 
$$
\left\{\begin{array}{l}y_{1}^{2}=6(2+t) \\ y_{2}^{2}=6(2-t)\end{array}\right.,
$$
and the coordinates of the midpoint of segment $AB$ are $\left(2, \frac{y_{1}+y_{2}}{2}\right)$. Note that $k_{AB}=\frac{y_{1}-y_{2}}{2 t}$, hence
$$
\frac{0 \frac{y_{1}+y_{2}}{2}}{x_{C}-2}=\frac{2 t}{y_{2}-y_{1}} \Rightarrow x_{C}=\frac{\left(y_{1}+y_{2}\right)\left(y_{1}-y_{2}\right)}{4 t}+2=\frac{y_{1}^{2}-y_{2}^{2}}{4 t}+2=\frac{6(2+t)-6(2-t)}{4 t}+2=5
$$

Thus, point $C$ is the fixed point $(5,0)$. Therefore, we have
$$
S_{\triangle ABC}=\frac{1}{2} \left| \begin{array}{ccc}
2+t & y_{1} & 1 \\
2-t & y_{2} & 1 \\
5 & 0 & 1
\end{array} \right| = \frac{1}{2} \left| t(y_{1}+y_{2}) + 3(y_{1}-y_{2}) \right|
$$

Since $12 t = y_{1}^{2} - y_{2}^{2} = (y_{1}+y_{2})(y_{1}-y_{2})$, we have $t(y_{1}+y_{2}) = \frac{(y_{1}+y_{2})^{2}(y_{1}-y_{2})}{12}$, hence
$$
\begin{array}{l}
S_{\triangle ABC} = \frac{1}{2} \left| \frac{(y_{1}+y_{2})^{2}(y_{1}-y_{2})}{12} + 3(y_{1}-y_{2}) \right| \\
= \frac{\left| (y_{1}+y_{2})^{2} + 36 \right| \cdot \sqrt{(y_{1}-y_{2})^{2}}}{24} = \frac{\left[ (y_{1}^{2} + y_{2}^{2}) + 36 + 2 y_{1} y_{2} \right] \cdot \sqrt{(y_{1}^{2} + y_{2}^{2}) - 2 y_{1} y_{2}}}{24} \\
= \frac{\left[ 60 + 2 y_{1} y_{2} \right] \cdot \sqrt{24 - 2 y_{1} y_{2}}}{24} = \frac{\sqrt{\left(30 + y_{1} y_{2}\right)^{2} \left(24 - 2 y_{1} y_{2}\right)}}{12} \leq \frac{\sqrt{\left(\frac{2(30 + y_{1} y_{2}) + (24 - 2 y_{1} y_{2})}{3}\right)^{3}}}{3} = \frac{14 \sqrt{7}}{12}
\end{array}
$$

The equality holds when and only when $y_{1} y_{2} = -2$. That is, when $A\left(\frac{6+\sqrt{35}}{3}, \sqrt{5}+\sqrt{7}\right), B\left(\frac{6-\sqrt{35}}{3}, \sqrt{5}-\sqrt{7}\right)$ or $A\left(\frac{6+\sqrt{35}}{3}, -\sqrt{5}-\sqrt{7}\right), B\left(\frac{6-\sqrt{35}}{3}, -\sqrt{5}+\sqrt{7}\right)$.