Since $P$ is the shared vertex between the three equilateral triangles, we note that $P$ is the incenter of $A B C$ since it is equidistant to all three sides. Since the area is 6 and the semiperimeter is also 6, we can calculate the inradius, i.e. the altitude, as 1, which in turn implies that the side length of the equilateral triangle is $\frac{2}{\sqrt{3}}$. Furthermore, since the incenter is the intersection of angle bisectors, it is easy to see that $A B_{2}=A C_{1}, B C_{2}=B A_{1}$, and $C A_{2}=C B_{1}$. Using the fact that the altitudes from $P$ to $A B$ and $C B$ form a square with the sides, we use the side lengths of the equilateral triangle to compute that $A B_{2}=A C_{1}=2-\frac{1}{\sqrt{3}}, B A_{1}=B C_{2}=1-\frac{1}{\sqrt{3}}$, and $C B_{1}=C A_{2}=3-\frac{1}{\sqrt{3}}$. We have that the area of the hexagon is therefore $$6-\left(\frac{1}{2}\left(2-\frac{1}{\sqrt{3}}\right)^{2} \cdot \frac{4}{5}+\frac{1}{2}\left(1-\frac{1}{\sqrt{3}}\right)^{2}+\frac{1}{2}\left(3-\frac{1}{\sqrt{3}}\right)^{2} \cdot \frac{3}{5}\right)=\frac{12+22 \sqrt{3}}{15}$$