1. Place the triangle conveniently using coordinates

Let $P=(0,0),\quad Q=(q,0),\quad R=(0,r).$ Then $\text{Area}(\triangle PQR)=\frac{1}{2}qr.$

2. Find the coordinates of points $A,B,C$ using the given ratios

Point $A$ on $QR$

Given $\frac{QA}{AR}=\frac{1}{2}.$ So $A$ divides $QR$ internally in the ratio $1:2$. Using section formula, $A=\left(\frac{2q+0}{3},\frac{2\cdot 0+r}{3}\right)=\left(\frac{2q}{3},\frac{r}{3}\right).$

Point $B$ on $RP$

Given $\frac{RB}{BP}=\frac{1}{2}.$ So $B$ divides $RP$ internally in the ratio $1:2$. Since $R=(0,r)$ and $P=(0,0)$, $B=\left(0,\frac{2r}{3}\right).$

Point $C$ on $PQ$

Given $\frac{PC}{CQ}=\frac{1}{2}.$ So $C$ divides $PQ$ internally in the ratio $1:2$. Hence $C=\left(\frac{q}{3},0\right).$

3. Find the area of $\triangle ABC$

Using the determinant formula, $\text{Area}(\triangle ABC)=\frac12\left|x_A(y_B-y_C)+x_B(y_C-y_A)+x_C(y_A-y_B)\right|.$

Substitute: $A\left(\frac{2q}{3},\frac{r}{3}\right),\quad B\left(0,\frac{2r}{3}\right),\quad C\left(\frac{q}{3},0\right).$

So, $\text{Area}(\triangle ABC)=\frac12\left|\frac{2q}{3}\left(\frac{2r}{3}-0\right)+0\left(0-\frac{r}{3}\right)+\frac{q}{3}\left(\frac{r}{3}-\frac{2r}{3}\right)\right|.$

=\frac12\left|\frac{4qr}{9}-\frac{qr}{9}\right|=rac12\cdot\frac{3qr}{9}=rac{qr}{6}.

4. Compute the required ratio