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` of incircle of the triangle XYZ is `(8pi)/3`

Options

A. (a) area of the triangle XYZ is `6sqrt6`
B. (b) the radius of circumcircle of the triangle XYZ is `(35)/(6) sqrt6`
C. (c) `sin.(X)/(2) sin.(Y)/(2) sin.(Z)/(2) = (4)/(35)`
D. (d) `sin^(2) ((X + Y)/(2)) = (3)/(5)`

Explanation

To solve the problem step by step, we will follow the given conditions and derive the necessary values.

### Step 1: Understand the given conditions
We have a triangle XYZ with sides opposite to angles X, Y, and Z denoted as x, y, and z respectively. The semi-perimeter \( s \) is given by:
\[ 2s = x + y + z \]
We also have the condition:
\[ \frac{s-x}{4} = \frac{s-y}{3} = \frac{s-z}{2} = \lambda \]
where \( \lambda \) is a constant.

### Step 2: Express the sides in terms of \( \lambda \)
From the equations, we can express \( s - x \), \( s - y \), and \( s - z \) in terms of \( \lambda \):
\[ s - x = 4\lambda \quad \Rightarrow \quad x = s - 4\lambda \]
\[ s - y = 3\lambda \quad \Rightarrow \quad y = s - 3\lambda \]
\[ s - z = 2\lambda \quad \Rightarrow \quad z = s - 2\lambda \]

### Step 3: Substitute into the semi-perimeter equation
Now, substituting \( x \), \( y \), and \( z \) into the semi-perimeter equation:
\[ x + y + z = (s - 4\lambda) + (s - 3\lambda) + (s - 2\lambda) = 3s - 9\lambda \]
Since \( 2s = x + y + z \):
\[ 2s = 3s - 9\lambda \]
Rearranging gives:
\[ s = 9\lambda \]

### Step 4: Find the lengths of the sides
Now that we have \( s \), we can find \( x \), \( y \), and \( z \):
\[ x = s - 4\lambda = 9\lambda - 4\lambda = 5\lambda \]
\[ y = s - 3\lambda = 9\lambda - 3\lambda = 6\lambda \]
\[ z = s - 2\lambda = 9\lambda - 2\lambda = 7\lambda \]

### Step 5: Find the area using Heron's formula
The area \( A \) of the triangle can be calculated using Heron's formula:
\[ A = \sqrt{s(s-x)(s-y)(s-z)} \]
Substituting the values:
\[ A = \sqrt{9\lambda \cdot (9\lambda - 5\lambda) \cdot (9\lambda - 6\lambda) \cdot (9\lambda - 7\lambda)} \]
\[ = \sqrt{9\lambda \cdot 4\lambda \cdot 3\lambda \cdot 2\lambda} = \sqrt{24 \cdot 9 \lambda^4} = 6\sqrt{6} \lambda^2 \]

### Step 6: Relate the area to the radius of the incircle
The area of the triangle can also be expressed in terms of the radius \( r \) of the incircle:
\[ A = r \cdot s \]
Given that the area is \( \frac{8\pi}{3} \):
\[ 6\sqrt{6} \lambda^2 = r \cdot 9\lambda \]
Thus:
\[ r = \frac{6\sqrt{6} \lambda^2}{9\lambda} = \frac{2\sqrt{6}}{3} \lambda \]

### Step 7: Solve for \( \lambda \)
Since the area is given as \( \frac{8\pi}{3} \), we can set up the equation:
\[ 6\sqrt{6} \lambda^2 = \frac{8\pi}{3} \]
Solving for \( \lambda^2 \):
\[ \lambda^2 = \frac{8\pi}{18\sqrt{6}} = \frac{4\pi}{9\sqrt{6}} \]

### Step 8: Final calculations
Now substituting back to find \( r \):
\[ r = \frac{2\sqrt{6}}{3} \cdot \sqrt{\frac{4\pi}{9\sqrt{6}}} = \frac{2\sqrt{6}}{3} \cdot \frac{2\sqrt{\pi}}{3\sqrt[4]{6}} = \frac{4\sqrt{\pi}}{9} \]

### Conclusion
The radius of the incircle and the area of the triangle have been calculated based on the given conditions.