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 s - x/4 = s - y/3 = s - z/2 and area of incircle of the triangle XYZ is 8π/3, then find the Area of the triangle XYZ, the radius of circumcircle of the triangle XYZ, sin²(X+Y)/2, and sinX/2sinY/2sinZ/2.

2S = x + y + z ⇒ s - x/4 = s - y/3 = s - z/2 = λ
S - x = 4λ
S - y = 3λ
S - z = 2λ

Adding all, we get S = 9λ, x = 5λ, y = 6λ, z = 7λ
πr² = 8π/3 ⇒ r² = 8/3
Δ = √S(S - x)(S - y)(S - z) ⇒ Δ = √9λ.4λ.3λ.2λ = 6√6λ²
R = xyz/4Δ = 5λ.6λ.7λ/4.6√6λ² = 35/(4√6)λ
we get λ = 1
(A) Δ = 6√6
(B) R = 35/(4√6)λ = 35/(4√6)
(C) r = 4RsinX/2sinY/2sinZ/2 ⇒ 2√(2/3) = 4.35/(4√6).sinX/2sinY/2sinZ/2
4/35 = sinX/2sinY/2sinZ/2
(D) sin²(X+Y)/2 = cos²Z/2 = S(S - z)/xy = 9.2/5.6 = 3/5