Area of Triangle

Let $[ABC]=k$ , let $[AXB]=x$ . Since $[ADB]=\frac{k}{3}, [ABE]=\frac{2k}{3}$ , we have $[BXD]=\frac{k}{3}-x, [AXE]=\frac{2k}{3}-x$ . Similarly, by the ratio of the bases (with the same height), we have $[CXE]=\frac{k}{3}-\frac{x}{2}, [DXC]=\frac{2k}{3}-2x$ . Adding all these up again we have $k=[ABC]=[AXB]+[BXD]+[AXE]+[CXE]+[DXC]=2k-\frac{7x}{2}$ . So $x=\frac{2k}{7}$ . By the same token, we also have $[AZC]=\frac{2k}{7}, [BYC]=\frac{2k}{7}$ . Therefore $[XYZ]=\frac{k}{7}=84$ . So $k=588$ .