In a `Delta ABC " if " |(1,a,b),(1,c,a),(1,b,c)| =0`, then `sin^(2) A + sin^(2) B + sin^(2) C` is

Options

A. `(3 sqrt3)/(2)`
B. `(9)/(4)`
C. `(5)/(4)`
D. 2

Explanation

To solve the problem, we need to analyze the determinant given and derive the required expression step by step.

### Step-by-Step Solution:

1. **Understanding the Determinant**:
We have the determinant:
\[ |(1,a,b),(1,c,a),(1,b,c)| = 0 \]
This determinant represents a condition on the angles of triangle ABC.

2. **Expanding the Determinant**:
The determinant can be expanded using the first column:
\[ |(1,a,b),(1,c,a),(1,b,c)| = 1 \cdot |(c,a),(b,c)| - 1 \cdot |(a,b),(b,c)| + 1 \cdot |(a,b),(c,a)| \]
This simplifies to:
\[ |(c,a),(b,c)| - |(a,b),(b,c)| + |(a,b),(c,a)| = 0 \]

3. **Calculating the 2x2 Determinants**:
Calculate each 2x2 determinant:
\[ |(c,a),(b,c)| = c \cdot c - a \cdot b = c^2 - ab \]
\[ |(a,b),(b,c)| = a \cdot c - b \cdot b = ac - b^2 \]
\[ |(a,b),(c,a)| = a \cdot a - b \cdot c = a^2 - bc \]

4. **Setting Up the Equation**:
Substitute these back into the determinant equation:
\[ (c^2 - ab) - (ac - b^2) + (a^2 - bc) = 0 \]
Rearranging gives:
\[ c^2 - ab - ac + b^2 + a^2 - bc = 0 \]

5. **Rearranging Terms**:
Combine like terms:
\[ a^2 + b^2 + c^2 - ab - ac - bc = 0 \]
This is the identity that holds for any triangle.

6. **Using the Identity**:
The identity \( a^2 + b^2 + c^2 - ab - ac - bc = 0 \) implies that the angles \( A, B, C \) are equal, leading to:
\[ A = B = C = \frac{\pi}{3} \]

7. **Finding \( \sin^2 A + \sin^2 B + \sin^2 C \)**:
Since \( A = B = C = \frac{\pi}{3} \):
\[ \sin^2 A = \sin^2 \left(\frac{\pi}{3}\right) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]
Thus,
\[ \sin^2 A + \sin^2 B + \sin^2 C = 3 \cdot \frac{3}{4} = \frac{9}{4} \]

### Final Answer:
\[ \sin^2 A + \sin^2 B + \sin^2 C = \frac{9}{4} \]