\(\triangle ABC\) has centroid \(G\).\(\triangle ABG\),\(triangle BCG\), and \(\triangle CAG\) have centroids \(G1\),\(G2\),\(G3\) respectively.we have to find out value of \(\frac{[G1G2G3]}{[ABC]}\) i.e area of \(\frac{[G1G2G3]}{[ABC]}\)

Let D, E, F be the midpoints of BC, CA, AB respectively.
Area of $\frac{[DEF]}{[ABC]}$=$\frac{1}{4}$

we know that any median is divided at the centroid $2:1$. Now can you find out \(GG_1,GG_2,GG_3\) ?

Can you now finish the problem ..........

we know that any median is divided at the centroid $2:1$
Now  $G_1$ is the centroid of $\triangle ABG$, then$GG_1=2G_1F$
Similarly,$GG_2 = 2G_2D$ and$GG_3 = 2G_3E$
Thus, From  homothetic transformation  $\triangle G_1G_2G_3$ maps to $\triangle FDE$ by a homothety of ratio$\frac{2}{3}$
Therefore,$\frac{[G_1G_2G_3]}{[DEF]}$ = $(\frac{2}{3})^2$=$\frac{4}{9}$

can you finish the problem........

Therefore we say that $\frac{[G_1G_2G_3]}{[ABC]}$ = $\frac{[G_1G_2G_3]}{[DEF]}\cdot \frac{[DEF]}{[ABC]} $= $\frac{4}{9}\cdot \frac{1}{4}$=$\frac{1}{9}$=$\frac{p}{q}$

So $p+q$=$9+1$=\(10\)