Let AX$$\bot$$BC of an equilateral triangle ABC. Then the sum of the perpendicular distances of the sides of $$\triangle$$ABC from any point inside the triangle is:

Let Z be a point inside the triangle as shown in the figure

ZD$$\bot $$BC,  ZE$$\bot $$AC,  ZF$$\bot $$AB and AB=BC=CA

Area of ($$\triangle $$ZAB + $$\triangle $$ZBC + $$\triangle $$ZCA) = Area of $$\triangle $$ABC

$$=$$>  $$\ \frac{1}{2}\times $$AB$$\times $$ZF + $$\ \frac{1}{2}\times $$BC$$\times $$ZD + $$\ \frac{1}{2}\times $$CA$$\times $$ZE = $$\ \frac{1}{2}\times $$BC$$\times $$AX

$$=$$>  $$\ \frac{1}{2}\times $$BC$$\times $$ZF + $$\ \frac{1}{2}\times $$BC$$\times $$ZD + $$\ \frac{1}{2}\times $$BC$$\times $$ZE = $$\ \frac{1}{2}\times $$BC$$\times $$AX

$$=$$>    ZF + ZD + ZE = AX

Hence, option A is the correct answer