Interesting Triangle Problem

On the sides $BC$ , $CA$ , $AB$ , of triangle $ABC$ , points $D$ , $E$ and $F$ , are taken in such a way that $\frac{BD}{DC}$ $=$ $\frac{CE}{EA}$ $=$ $\frac{AF}{FB}$ $=$ $\frac{2}{1}$ . If area of triangle $ABC$ is $106.61$ square unit, then the area of the triangle enclosed by the lines $AD$ , $BE$ and $CF$ , i.e. area of triangle $PQR$ is

Note: Figure is not to scale