Let P Q R   be a triangle. The points A ,   B and C are on the sides Q R ,   R P and P Q…

Let the position vector of $P$ is $\overrightarrow{O}$, $Q$ is $\overrightarrow{q}$ and $R$ is $\overrightarrow{r}$.

Now, position vector of $A$ is

$\frac{2\overrightarrow{q}+\overrightarrow{r}}{3}$

Position vector of $B$ is $\frac{2\overrightarrow{r}}{3}$.

Position vector of $C$ is $\frac{\overrightarrow{q}}{3}$.

$\overrightarrow{AB}=\frac{\overrightarrow{r}-2\overrightarrow{q}}{3}$

$\overrightarrow{AC}=\frac{-\overrightarrow{r}-\overrightarrow{q}}{3}$

So,

$\overrightarrow{AB}\times \overrightarrow{AC}=\frac{1}{3}\left[\left(\overrightarrow{r}-2\overrightarrow{q}\right)\times \left(-\overrightarrow{r}-\overrightarrow{q}\right)\right]$

$\Rightarrow \overrightarrow{AB}\times \overrightarrow{AC}=\frac{1}{3}\left[-\left(\overrightarrow{r}\times \overrightarrow{q}\right)+2\left(\overrightarrow{q}\times \overrightarrow{r}\right)\right]$

$\Rightarrow \overrightarrow{AB}\times \overrightarrow{AC}=\left(\overrightarrow{q}\times \overrightarrow{r}\right)$

Area of $∆PQR$ is $=\frac{1}{2}\left|\overrightarrow{q}\times \overrightarrow{r}\right|$

Area of $∆ABC$ is $\frac{1}{2}\left|\overrightarrow{AB}\times \overrightarrow{AC}\right|$

Area of $∆ABC=\frac{1}{2}\left|\overrightarrow{q}\times \overrightarrow{r}\right|$

Therefore,

$\frac{\mathrm{Area} \left(∆PQR\right)}{\mathrm{Area}\left(∆ABC\right)}=1$