In a triangle 'X:YZ, let x, y, z be the lengths of sides opposite to the angles X, Y. Z respectively, and $2s=x+y+Z$

If $\frac{s-x}{4}=\frac{s-y}{3}=\frac{s-z}{2}$ and area of incircle of the triangle xyz is $\frac{8\pi }{3}$, then 

Let $\frac{s-x}{4}=\frac{s-y}{3}=\frac{s-z}{2}=k$

$\Rightarrow s-x=4k,s-y=3k,s-z=2k$

on adding, we get $s=9k$

$\therefore x=5k,y=6k,z=7k$

$\begin{array}{l}{\mathrm{\Delta }}^2=\left(9\mathrm{K}\right)\left(4\mathrm{K}\right)\left(3\mathrm{K}\right)\left(2\mathrm{k}\right)\\ \therefore \mathrm{\Delta }=6\sqrt{6}{\mathrm{K}}^2\end{array}$

Given that $\pi r^2=\frac{8\pi }{3}$

$\Rightarrow {\left(\frac{\mathrm{\Delta }}{S}\right)}^2=\frac{8}{3}\Rightarrow \frac{6\sqrt{6}{K}^2}{9K}=\frac{\sqrt{8}}{\sqrt{3}}\Rightarrow K=1$

$\text{(A) }\mathrm{\Delta }=6\sqrt{6}$

$\text{(B) }R=\frac{xyz}{4\mathrm{\Delta }}=\frac{\mathrm{5.6.7}}{24\sqrt{6}}=\frac{35}{4\sqrt{6}}$

(C) We have $r=\frac{2\sqrt{2}}{\sqrt{3}}$

$\begin{array}{l}\Rightarrow 4\cdot \frac{35}{4\sqrt{6}}\sin \frac{x}{2}\sin \frac{y}{2}\sin \frac{z}{2}=\frac{2\sqrt{2}}{\sqrt{3}}\\ \Rightarrow \sin \frac{x}{2}\sin \frac{y}{2}\sin \frac{z}{2}=\frac{4}{35}\end{array}$

$\text{(D) }{\sin }^2\frac{x+y}{2}={\cos }^2\frac{z}{2}=\frac{s(s-z)}{xy}=\frac{9.2}{5.6}=\frac{3}{5}$