Right angled triangle 
My approach: Take $B$ as origin and $C$as $(15,0)$ and $P$ as $(6,0)$ and $G$ as $(6,b)$ and $A$ as $(3,3b)$.
So we have all the vertices of the triangle except the $y$-coordinate of one vertex,which is three times the y-coordinate of the centroid.
What to do next ?
If i write the equation , i get everything in terms of the variable $b$ and i don't see how to eliminate it.
I see that the small right angled triangle and larger triangle are similar but does that help ?
  Can we do it without coordinate geometry ?

Let's continue with your method. Drop a perpendicular from $A$ of the line $BC$ and call the point $M$. Using similar triangles (which does help), we have:
$$\frac{BM}{AM} = \frac{AM}{CM}$$
$$\frac{3}{3b} = \frac{3b}{12}$$
$$9b^2=36 ,b=2$$
which explains where the formula $h^2=pq$ comes from.
Using similar triangles again, we have:
$$\frac{AM}{CM} = \frac{QP}{PC}$$
$$\frac{3 \cdot 2}{ 12} = \frac{QP}{9}$$
and you can surely continue from here.
