### Instruction:
Point $P$ is inside equilateral triangle $ABC$ such that the altitudes from $P$ to $\overline{AB}$, $\overline{BC}$, and $\overline{CA}$ have lengths 5, 6, and 7 respectively.  What is the area of triangle $ABC$?

### Response:

We begin by drawing a diagram:

[asy]
pair A,B,C,P,X,Y,Z;
real s=12*sqrt(3);
A=(0,0); C=(s,0); B=(s/2,s/2*sqrt(3)); P=(9.5,7); X= foot(P,B,C); Y=foot(P,A,B); Z=foot(P,A,C);
draw(A--B--C--cycle); draw(P--Z); draw(P--Y); draw(P--X);
draw(rightanglemark(P,X,B,25)); draw(rightanglemark(P,Z,C,25)); draw(rightanglemark(P,Y,A,25));

label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$P$",P,SE);
label("$7$",P--Z,W); label("$6$",P--X,S); label("$5$",P--Y,NE);
[/asy]

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: $\triangle APB$, $\triangle BPC$, and $\triangle CPA$.

[asy]
pair A,B,C,P,X,Y,Z;
real s=12*sqrt(3);
A=(0,0); C=(s,0); B=(s/2,s/2*sqrt(3)); P=(9.5,7); X= foot(P,B,C); Y=foot(P,A,B); Z=foot(P,A,C);

label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$P$",P,SE);
label("$7$",P--Z,W); label("$6$",P--X,S); label("$5$",P--Y,NE);

fill(P--A--B--cycle,rgb(135,206,250));

fill(P--A--C--cycle,yellow);
fill(P--B--C--cycle,rgb(107,142,35));

draw(P--A,dashed); draw(P--B,dashed); draw(P--C,dashed);
draw(A--B--C--cycle); draw(P--Z); draw(P--Y); draw(P--X);

[/asy]

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

We have \[[\triangle ABC] = [\triangle APB] + [\triangle BPC] + [\triangle CPA],\]or  \begin{align*}
\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{align*}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 \[[\triangle ABC] = \frac{s^2\sqrt{3}}{4}=\left(\frac{s\sqrt{3}}{4}\right)(s) = (9)(12\sqrt{3})=108\sqrt{3}.\]