Claim: $G$ is the centroid of $\triangle PQR$. We have $A(x_1, y_1), B(x_2, y_2), C(x_3, y_3)$ and $G(\tfrac{x_1+x_2+x_3}{3}, \tfrac{y_1+y_2+y_3}{3})$. $P(\tfrac{x_2-x_1}{k}+x_1, \tfrac{y_2-y_1}{k}+y_1)$ $Q(\tfrac{x_3-x_2}{k}+x_2, \tfrac{y_3-y_2}{k}+y_2)$ $P(\tfrac{x_1-x_3}{k}+x_3, \tfrac{y_1-y_3}{k}+y_3)$. $G'(\tfrac{x_1+x_2+x_3}{3}, \tfrac{y_1+y_2+y_3}{3})$. So $G'=G$ is the centroid of $\triangle PQR$. Thus the desired ratio is $\frac13$

From here, $G$ is the centroid of $PQR$ and so, $\frac{[PQG]}{[PQR]}=\frac{1}{3}$.