Question 51
Convex hexagon $A B C D E F$ is drawn in the plane such that $A C D F$ and $A B D E$ are parallelograms with area $168$. $A C$ and $B D$ intersect at $G$. Given that the area of $A G B$ is $10$ more than the area of $C G B$, find the smallest possible area of hexagon $A B C D E F$.