Let the midpoint of $AB$ be $M(x_0, y_0)$. Then $x_0 = \frac{x_1 + x_2}{2} = 2$ and $y_0 = \frac{y_1 + y_2}{2}$. We have
$$k_{AB} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\frac{y_2}{6} - \frac{y_1}{6}}{\frac{y_2}{6} + \frac{y_1}{6}} = \frac{\frac{6}{y_2} + \frac{6}{y_1}}{\frac{6}{y_2} + \frac{6}{y_1}} = \frac{3}{y_0}.$$
The equation of the perpendicular bisector of $AB$ is
$$y - y_0 = -\frac{y_0}{3}(x - 2). \qquad \textcircled{1}$$
It is easy to find that one solution of it is $x = 5$, $y = 0$.
Therefore, the intersection $C$ is a fixed point with coordinate $(5, 0)$.
From ①, we know the equation of line $AB$ is $y - y_0 = \frac{3}{y_0}(x - 2)$, or
$$x = \frac{y_0}{3}(y - y_0) + 2. \qquad \textcircled{2}$$
Substituting ② in $y^2 = 6x$, we get $y^2 = 2y_0(y - y_0) + 12$, or
$$y^2 - 2y_0y + 2y_0^2 - 12 = 0. \qquad \textcircled{3}$$
As $y_1$ and $y_2$ are two real roots of ③ and $y_1 \neq y_2$, we have
$$\Delta = 4y_0^2 - 4(2y_0^2 - 12) = -4y_0^2 + 48 > 0.$$
Therefore, $-2\sqrt{3} < y_0 < 2\sqrt{3}$. Then we have

The distance from point $C(5, 0)$ to segment $AB$ is
$$h = |CM| = \sqrt{(5-2)^2 + (0-y_0)^2} = \sqrt{9+y_0^2}.$$
Therefore,
$$\begin{align*} S_{\triangle ABC} &= \frac{1}{2} |AB| \cdot h = \frac{1}{3} \sqrt{(9+y_0^2)(12-y_0^2)} \cdot \sqrt{9+y_0^2} \\ &= \frac{1}{3} \sqrt{\frac{1}{2}(9+y_0^2)(24-2y_0^2)(9+y_0^2)} \\ &\le \frac{1}{3} \sqrt{\frac{1}{2} \left( \frac{9+y_0^2+24-2y_0^2+9+y_0^2}{3} \right)^3} \\ &= \frac{14}{3}\sqrt{7}. \end{align*}$$
The equality holds if and only if $9 + y_0^2 = 24 - 2y_0^2$, i.e.
$$y_0 = \pm\sqrt{5}. \text{ Then we get}$$
$$\text{and} \quad A\left(\frac{6+\sqrt{35}}{3}, \sqrt{5}+\sqrt{7}\right), B\left(\frac{6-\sqrt{35}}{3}, \sqrt{5}-\sqrt{7}\right)$$
$$A\left(\frac{6+\sqrt{35}}{3}, - (\sqrt{5}+\sqrt{7})\right), B\left(\frac{6-\sqrt{35}}{3}, -\sqrt{5}+\sqrt{7}\right).$$

Similar to Solution 1, we get that $C$, the intersection of the perpendicular bisector of $AB$ and the $x$-axis, is a fixed point with coordinate $(5, 0)$.
Let $x_1 = t_1^2$, $x_2 = t_2^2$, $t_1 > t_2$, $t_1^2 + t_2^2 = 4$. Then $S_{\triangle ABC}$ is the absolute value of
$$\frac{1}{2} \begin{vmatrix} 5 & 0 & 1 \\ t_1^2 & \sqrt{6} t_1 & 1 \\ t_2^2 & \sqrt{6} t_2 & 1 \end{vmatrix},$$
so
$$\begin{aligned} S_{\triangle ABC}^2 &= \left( \frac{1}{2} (5\sqrt{6} t_1 + \sqrt{6} t_1^2 t_2 - \sqrt{6} t_1 t_2^2 - 5\sqrt{6} t_2) \right)^2 \\ &= \frac{3}{2} (t_1 - t_2)^2 (t_1 t_2 + 5)^2 \\ &= \frac{3}{2} (4 - 2t_1t_2)(t_1t_2 + 5)(t_1t_2 + 5) \\ &\le \frac{3}{2} \left( \frac{14}{3} \right)^3. \end{aligned}$$
Therefore, $S_{\triangle ABC} \le \frac{14}{3}\sqrt{7}$ and the equality holds if and only if $(t_1 - t_2)^2 = t_1t_2 + 5$ and $t_1^2 + t_2^2 = 4$. We then get $t_1 = \frac{\sqrt{7} + \sqrt{5}}{\sqrt{6}}$ and $t_2 = -\frac{\sqrt{7} - \sqrt{5}}{\sqrt{6}}$, which implies either
$$A\left(\frac{6 + \sqrt{35}}{3}, \sqrt{5} + \sqrt{7}\right), B\left(\frac{6 - \sqrt{35}}{3}, \sqrt{5} - \sqrt{7}\right)$$
or
$$A\left(\frac{6 + \sqrt{35}}{3}, - (\sqrt{5} + \sqrt{7})\right), B\left(\frac{6 - \sqrt{35}}{3}, -\sqrt{5} + \sqrt{7}\right).$$