P is a point in the interior of an equilateral $$\triangle ABC$$. Line-segments PD, PE, PF are the perpendiculars drawn from P to the sides BC, CA, AB respectively with D lies on side BC, E lies on side CA and F lies on side AB. If $$PD = 6$$ cm, $$PE = 8$$ cm, $$PF = 10$$ cm, then area of the triangle ABC is _____ $$cm^2$$

By Viviani's theorem the three perpendicular distances add up to the altitude, so $$h = 6 + 8 + 10 = 24$$ cm. For an equilateral triangle of side $$a$$, $$h = \frac{\sqrt{3}}{2}a$$, so $$a = \frac{48}{\sqrt{3}} = 16\sqrt{3}$$ cm. The area is $$\frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{4} \times 768 = 192\sqrt{3}$$ $$\text{cm}^2$$, which is about $$332.55$$ $$\text{cm}^2$$.