Your idea matters-3

$[\Delta OPR] = \frac{5 \cdot 4}{2} = 10$

$[\Delta OPQ] : [\Delta OQR] :: 3 : 2$

$\therefore [\Delta OQR] = \frac{2}{5} \cdot 10 \implies \boxed{4}$

The $x$ -coordinate of the point $Q$ is $\dfrac{2\times 5+3\times0}{2+3}=2$ , which is the height of the $\triangle OQR$ . Therefore area of $\triangle OQR$ is $\dfrac{1}{2}\times 2\times 4=\boxed{4}$

Here,

$\triangle OPR=\frac {1}{2}×5×4=10$

We know , if the hight of two triangles remains same, the area of them will be proportional to their base.

So, $\triangle OQR=\frac{2}{5}×10=\boxed {4}$