This problem is from 2019 Math Kangaroo competition for 9th-10th graders that took place last week, problem #29.
I was able to solve it using coordinate geometry, both triangles have the same area. However, I do not expect 9th graders to know this method. Is there a simpler solution that I am not seeing?
Best Answer
Since $D$ is the midpoint of $BC$, $A_\triangle ACD=A_\triangle ABD=\frac{1}{2}S$.
Since $AP=2AB$ and $AQ=3AD$, $A_\triangle APQ$ is $2\times 3=6$ times $A_\triangle ABD$. Similarly $A_\triangle AQR$ and $A_\triangle APR$. So $A_\triangle PQR = A_\triangle APQ+A_\triangle AQR - A_\triangle APR$, giving the answer.
All this is just the ratio of areas of triangle with same base and ratio of height (or vice versa), which a year 9 student should already know.