Let $A^{\prime}$ be the point on $BC$ such that $2BA^{\prime}=A^{\prime}C$. By law of cosines on triangle $AA^{\prime}B$, we find that $AA^{\prime}=2\sqrt{7}$. By power of a point, $A^{\prime}A_{1}=\frac{2 \cdot 4}{2\sqrt{7}}=\frac{4}{\sqrt{7}}$. Using side length ratios, $A_{1}A_{2}=2\frac{AA_{1}}{AA^{\prime}}=2\frac{2\sqrt{7}+\frac{4}{\sqrt{7}}}{2\sqrt{7}}=\frac{18}{7}$. Now our hexagon can be broken down into equilateral triangle $A_{1}B_{1}C_{1}$ and three copies of triangle $A_{1}C_{1}C_{2}$. Since our hexagon has rotational symmetry, $\angle C_{2}=120$, and using law of cosines on this triangle with side lengths $\frac{18}{7}$ and 6, a little algebra yields $A_{1}C_{2}=\frac{30}{7}$ (this is a 3-5-7 triangle with an angle 120). The area of the hexagon is therefore $\frac{6^{2}\sqrt{3}}{4}+3 \cdot \frac{1}{2} \cdot \frac{18}{7} \cdot \frac{30}{7} \cdot \frac{\sqrt{3}}{2}=\frac{846\sqrt{3}}{49}$.