10. (20 points) Given that two vertices of an equilateral triangle lie on the parabola $y^{2}=4 x$, the third vertex lies on the directrix of the parabola, and the distance from the center of the triangle to the directrix is equal to $\frac{1}{9}$ of the perimeter. Find the area of the triangle.
10. As shown in Figure 3, let the focus be $F$, the centroid of $\triangle A B C$ be $G$, and the projections of points $A$, $B$, and $G$ on the directrix be $A_{1}$, $B_{1}$, and $G_{1}$, respectively. According to the coordinates of the centroid, we have
$$
\left|G G_{1}\right|=\frac{1}{3}\left(\left|A A_{1}\right|+\left|B B_{1}\right|\right) \text {. }
$$

From the problem, we know
$$
\begin{array}{l}
\left|G G_{1}\right|=\frac{1}{9}(|A B|+|B C|+|C A|)=\frac{1}{3}|A B| \\
\Rightarrow|A B|=\left|A A_{1}\right|+\left|B B_{1}\right|=|A F|+|F B| \\
\Rightarrow A, F, B \text { are collinear. }
\end{array}
$$

Taking the midpoint $M$ of $A B$, and let the projection of $M$ on the directrix be $M_{1}$. We have
$$
\begin{array}{l}
\text { By }|M C|=\frac{\sqrt{3}}{2}|A B|, \\
\left|M M_{1}\right|=\frac{1}{2}\left(\left|A A_{1}\right|+\left|B B_{1}\right|\right)=\frac{1}{2}|A B|,
\end{array}
$$

Thus, $|M C|=\sqrt{3}\left|M M_{1}\right|$
$$
\Rightarrow\left|M_{1} C\right|=\sqrt{2}\left|M M_{1}\right| \text {, }
$$

which means the slope of line $M C$ is $\pm \sqrt{2}$, and the slope of line $A B$ is $\pm \frac{\sqrt{2}}{2}$.
Therefore, the equation of line $A B$ is $y= \pm \frac{\sqrt{2}}{2}(x-1)$.
Combining with the equation of the parabola, we find $|A B|=12$.
Thus, the area of $\triangle A B C$ is $36 \sqrt{3}$.