Let $PQR$  be a triangle. The points $A,B$ and $C$ are on the sides $QR,RP$ and $PQ$ respectively such that $\frac{QA}{AR}=\frac{RB}{BP}=\frac{PC}{CQ}=\frac{1}{2}$. Then $3\frac{Area\left(\Delta PQR\right)}{Area\left(\Delta ABC\right)}$ is equal to

Let P is $\overrightarrow{0},Q$  is $\overrightarrow{q}$ and $R$  is  $\overrightarrow{r},$
A is $\frac{2\overrightarrow{q}+\overrightarrow{r}}{3},B$  is $\frac{2\overrightarrow{r}}{3}$  and C is  $\frac{\overrightarrow{q}}{3}$
Since, area of  $\Delta PQR=\frac{1}{2}|\overrightarrow{q}\times \overrightarrow{r}|$
Area of  $\Delta ABC=\frac{1}{2}|\overline{AB}\times \overline{AC}|$
As $\overline{AB}=\frac{\overrightarrow{r}-2\overrightarrow{q}}{3},\overline{AC}=\frac{-\overrightarrow{r}-\overrightarrow{q}}{3}$  
So, Area of  $\Delta ABC=\frac{1}{6}|\overrightarrow{q}\times \overrightarrow{r}|$
Now, $\frac{Area\left(\Delta PQR\right)}{Area\left(\Delta ABC\right)}=3$