smc

Let $O_1$ and $O_2$ be the centers of ${\omega }_1$ and ${\omega }_2$ respectively and draw $O_1{O}_2$, $O_1{P}_1$, and $O_2{P}_2$. Note that $\angle O_1{P}_1{P}_2$ and $\angle O_2{P}_2{P}_3$ are both right. Furthermore, since $△P_1{P}_2{P}_3$ is equilateral, $m\angle P_1{P}_2{P}_3=60^{\circ }$ and $m\angle O_2{P}_2{P}_1=30^{\circ }$. Mark $M$ as the base of the altitude from $O_2$ to $P_1{P}_2.$ Since $△P_2{O}_{2}M$ is a 30-60-90 triangle, $O_{2}M=2$ and $P_{2}M=2\sqrt{3}$. Also, since $O_1{O}_2=8$ and $O_1{P}_1=4$, we can find $P_{1}M=\sqrt{8^2-(4+2{)}^2}=2\sqrt{7}$. Thus, $P_1{P}_2=P_{1}M+P_{2}M=2\sqrt{3}+2\sqrt{7}$. This makes $$\left[P_1{P}_2{P}_3\right]=\frac{\sqrt{3}}{4}\cdot {\left(2\sqrt{3}+2\sqrt{7}\right)}^2=10\sqrt{3}+6\sqrt{7}=\sqrt{300}+\sqrt{252}.$$ So, our answer is $252+300=\boxed{552}$. Note by diyarv: A way to see that $P_{1}M=2\sqrt{7}$ is through similar triangles. Call the intersection between $P_{1}M$ and $O_1{O}_2$ $S$. Since angles $\angle O_1{P}_{1}S$ and $\angle O_{2}MS$ are both right angles, and $\angle P_{1}S{O}_1$ and $\angle MS{O}_2$ are congruent, triangles $△P_{1}S{O}_1$ and $△MS{O}_2$ are similar by AA similarity. And, since $O_1{P}_1$ is $4$ and $O_{2}M$ is 2, the common ratio is $2$. Using the fact that $O_{1}S+S{O}_2=8$, we see that these lengths are $\frac{16}{3}$ and $\frac{8}{3}$ respectively. Using the Pythagorean theorem, we see that $P_{1}S$ and $S{P}_2$ are $\frac{2\sqrt{7}}{3}$ and $\frac{4\sqrt{7}}{3}$ respectively, giving us a sum of $2\sqrt{7}$.