In a triangle XYZ, let x, y, z be the lengths of sides opposite to the angles X, Y, Z, respectively, and $$2s = x + y + z$$. If $$\frac{s - x}{4} = \frac{s - y}{3} = \frac{s - z}{2}$$ andareaofincircle of the triangle XYZ is $$\frac{8 \pi}{3}$$, Then

A

area of the triangle XYZ is $$6\sqrt{6}$$

B

the radiusof circumcircle of the triangle XYZ is $$\frac{35}{6}\sqrt{6}$$

C

$$\sin\frac{X}{2}\sin\frac{Y}{2}\sin\frac{Z}{2} = \frac{4}{35}$$

D

$$\sin^2\left(\frac{X + Y}{2}\right) = \frac{3}{5}$$