jmc

We begin by drawing a diagram:



Let the side length of triangle $ABC$ be $s$; since it is equilateral, its area is $\frac{s^2\sqrt{3}}{4}$.

Now, we draw segments from $P$ to the three vertices of triangle $ABC$, which divides the triangle into three smaller triangles: $△APB$, $△BPC$, and $△CPA$.



We can compute the area of these three small triangles, and sum their areas to get the area of equilateral $△ABC$. We compute the area of triangle $APB$ by using $AB$ as the base and 5 as the height. $AB$ has length $s$, so $$[△APB]=\frac{1}{2}(s)(5).$$Similarly, $[△BPC]=\frac{1}{2}(s)(6)$ and $[△APC]=\frac{1}{2}(s)(7)$.

We have $$[△ABC]=[△APB]+[△BPC]+[△CPA],$$or $$\begin{array}{rl}\frac{s^2\sqrt{3}}{4}& =\frac{1}{2}(s)(5)+\frac{1}{2}(s)(6)+\frac{1}{2}(s)(7)\\ & =\frac{1}{2}(s)(5+6+7)\\ & =9s.\end{array}$$We can divide both sides of the above simplified equation by $s$, since side lengths are positive and not zero, to get $\frac{s\sqrt{3}}{4}=9$. Solving for $s$ gives $$s=9\cdot \frac{4}{\sqrt{3}}=12\sqrt{3}.$$Finally, the area of triangle $ABC$ is $$[△ABC]=\frac{s^2\sqrt{3}}{4}=\left(\frac{s\sqrt{3}}{4}\right)(s)=(9)(12\sqrt{3})=\boxed{108\sqrt{3}}.$$