Let $A X$ and $B D$ meet at $P$. We have $D P / P B=D X / A B=3 / 4$. Now, applying Ceva's Theorem in triangle $A B D$, we see that $$\frac{A Z}{Z B}=\frac{D P}{P B} \cdot \frac{A Y}{Y D}=\frac{3}{4} \cdot \frac{1}{2}=\frac{3}{8}$$ Now, $$\frac{[A Y Z]}{[A B C D]}=\frac{[A Y Z]}{2[A B D]}=\frac{1}{2} \cdot \frac{1}{3} \cdot \frac{3}{11}=\frac{1}{22}$$ and similarly $$\frac{[D Y X]}{[A B C D]}=\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4}=\frac{1}{4}$$ Also, $$\frac{[X C B Z]}{[A B C D]}=\frac{1}{2}\left(\frac{1}{4}+\frac{8}{11}\right)=\frac{43}{88}$$ The area of $X Y Z$ is the rest of the fraction of the area of $A B C D$ not covered by the three above polygons, which by a straightforward calculation $19 / 88$ the area of $A B C D$, so our answer is $$8 \cdot 11 \cdot \sin 60^{\circ} \cdot \frac{19}{88}=\frac{19 \sqrt{3}}{2}$$