Given a point $P$ outside equilateral $\Delta ABC$ but inside $\angle ABC$, if the distance between $P$ to $BC,CA,AB$ are $h_1,h_2,h_3$ respectively. 
Given a point $P$ outside equilateral $\Delta ABC$ but inside $\angle ABC$, if the distance between $P$ to $BC,CA,AB$ are $h_1,h_2,h_3$ respectively, where $h_1 - h_2 + h_3 = 6$, find $[\Delta ABC]$ .

What I Tried: At first I couldn't understand if $h_1,h_2,h_3$ are just any lines touching the sides or are some specific lines like altitudes or medians (bisecting the sides of the triangle) . But since they are denoted like $h_1,h_2,h_3$ I suppose they are the altitudes. So here is a picture :-

No idea for this problem. I don't think I can use any simple geometry techniques here like angle-chasing, area of triangles, pythagorean theorem and so on, because I have been given very less info. So I am a bit stuck here.
Can anyone help me? Thank You.
 A: Connect point $P$ to $A, B, C$. Now you can see that $\triangle ABC = \triangle PAB + \triangle PBC - \triangle PAC = \frac{AB}{2}(h_1 - h_2 + h_3) = 3 AB$
Given equilateral triangle, we also know that $\triangle ABC = \frac{\sqrt3}{4}AB^2 = 3 AB$
That gives you the value of $AB = 4 \sqrt3$ and $\triangle ABC = 12 \sqrt3$
A: Consider the areas of triangles $APB, BPC, CPA$.
We have the equation
\begin{align}
[\triangle ABC] &= [\triangle APB] - [\triangle CPA] + [\triangle BPC]\\
&=\frac12(ABh_3 - AC h_2 + BC h_1)\\
&=\frac{AB}2(h_1-h_2+h_3)\\
&=3AB
\end{align}
We also have the relation between the side of an equilateral triangle and its area:
$$[\triangle ABC] = \frac{\sqrt 3}4AB^2$$
Now solve for $AB$ and $[\triangle ABC]$.
A: By Viviani's theorem, the height of $\triangle ABC$ is $6$, so its side length is $\frac{12}{\sqrt3}=4\sqrt3$ and its area is $\frac{6×4\sqrt3}2=12\sqrt3$.
A: WLOG let $y=0,y=\sqrt{3}x,y=-\sqrt{3}(x-a)$ be the  eq of  sides  triangle with side of equilateral being $a$.
thus $$h_1+h_3-h_2=1$$ $$|k+\frac{\sqrt{3}h-k}{2}-\frac{\sqrt{3}h+k +\sqrt{3}a}{2}|=1$$
$$\sqrt{3}a/2=1$$
$$a=?$$
$$area(\Delta)={\sqrt{3}a^2}/4=?$$
