On the sides $BC$, $CA$, $AB$ of $\Delta ABC$, points $D$, $E$, $F$ are taken in such a way that $\frac{BD}{DC} = \frac{CE}{EA} = $ How to find the area ?
The Diagram and the question.

Let $G\in EC$ such that $DG||BE$ and $GC=2x$.
Thus, by Thales $$\frac{EG}{2x}=\frac{BD}{DC}=2,$$ which gives $EG=4x$, $EC=6x$ and $AE=3.$
Thus, by Thales again we obtain:
$$\frac{AP}{PD}=\frac{AE}{EG}=\frac{3x}{4x}=\frac{3}{4}.$$
Now, let $I\in DC$ such that $EI||AD$ and $DI=y$.
Thus, by Thales $IC=2y$ and from here $DC=3y$ and $BD=6y$ and by Thales again we obtain:
$$\frac{BP}{PE}=\frac{BD}{DI}=\frac{6y}{y}=6.$$
Id est, $$AP:PR:RD=BQ:QP:PE=CR:RQ:QF=3:3:1.$$
Hence, $$S_{\Delta PQR}=\frac{1}{2}S_{\Delta QPC}=\frac{1}{2}\cdot\frac{3}{4}S_{\Delta QEC}=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{4}{7}S_{\Delta BEC}=$$
$$=\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{4}{7}\cdot\frac{2}{3}S_{\Delta ABC}=\frac{1}{7}S_{\Delta ABC}=\frac{1}{7}\cdot106.61=15.23.$$
