I5.3 Given that the area of the $\triangle A B C$ is $3 Q ; D, E$ and $F$ are the points on $A B, B C$ and $C A$ respectively such that $A D=\frac{1}{3} A B, B E=\frac{1}{3} B C, C F=\frac{1}{3} C A$. If the area of $\triangle D E F$ is $R$, find the value of $R$.
$\begin{array}{l}R=3-\text { area } \triangle A D F-\text { area } \triangle B D E-\text { area } \triangle C E F \\ =3-\left(\frac{1}{2} A D \cdot A F \sin A+\frac{1}{2} B E \cdot B D \sin B+\frac{1}{2} C E \cdot C F \sin C\right) \\ =3-\frac{1}{2}\left(\frac{c}{3} \cdot \frac{2 b}{3} \sin A+\frac{2 c}{3} \cdot \frac{a}{3} \sin B+\frac{2 a}{3} \cdot \frac{b}{3} \sin C\right) \\ =3-\frac{2}{9}\left(\frac{1}{2} \cdot b c \sin A+\frac{1}{2} \cdot a c \sin B+\frac{1}{2} \cdot a b \sin C\right) \\ =3-\frac{2}{9}(3 \times \text { area of } \triangle \mathrm{ABC}) \\ =3-\frac{2}{9} \times 9=1\end{array}$