Let $P$ be arbitrary point inside the triangle $ABC$. The points $K$, $L$ and $M$ are points of intersection of the perpendicular lines passing through the point $P$ to the sides $AB$, $BC$ and $CA$ respectively. Since
$$\angle PKB = \angle PLB = \angle PLC = \angle PMC = \angle PMA = \angle AKP = 90^\circ,$$
the quadrilaterals $AKPM$, $BLPK$ and $CLPM$ are cyclic quadrilaterals. From this we get $\angle KPL = \angle LPM = \angle MPK = 120^\circ$ because the angles in the triangle $ABC$ are equal to $60^\circ$. Since $\overline{PK} = p$, $\overline{PL} = m$ and $\overline{PM} = n$ we obtain
$$P_{\angle LPK} = \frac{1}{2} mp \sin 120^\circ = \frac{\sqrt{3}}{4} mp,$$
$$P_{\angle LPM} = \frac{1}{2} mn \sin 120^\circ = \frac{\sqrt{3}}{4} mn,$$
$$P_{\angle MPK} = \frac{1}{2} np \sin 120^\circ = \frac{\sqrt{3}}{4} np.$$
The area of the triangle $KLM$ is equal to
$$P_{\Delta KLM} = P_{\Delta KPL} + P_{\Delta LPM} + P_{\Delta MPK} = \frac{\sqrt{3}}{4} (pm + mn + np)$$
Because the area of the triangle $ABC$ is
$$P_{\Delta ABC} = \frac{\sqrt{3}}{4} a^2, \text{ we obtain } \frac{P_{\Delta ABC}}{P_{\Delta KLM}} = \frac{\frac{\sqrt{3}}{4} a^2}{\frac{\sqrt{3}}{4} (pm + mn + np)} = \frac{a^2}{pm + mn + np}.$$