Give me some hint.
Where is the point “P” ?
Let G be the intersection of AD and FC. From vertical angles, we know that \angle FGA= \angle DGC. Also, because we are given that ABCD and AFCE are rectangles, we know that \angle AFG= \angle CDG=90 ^{\circ}. Therefore, by AA similarity, we know that \triangle AFG\sim\triangle CDG.Let AG=x. Then, we have DG=11-x. By similar triangles, we know that FG=\frac{7}{3}(11-x) and CG=\frac{3}{7}x. We have \frac{7}{3}(11-x)+\frac{3}{7}x=FC=9.