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 necessary conditions for the triangle ABC.

### Step-by-Step Solution:

1. **Understanding the Determinant**: We are given the determinant:
\[ D = \begin{vmatrix} 1 & a & b \\ 1 & c & a \\ 1 & b & c \end{vmatrix} = 0 \]
This determinant being zero implies that the rows are linearly dependent.

2. **Expanding the Determinant**: We can expand this determinant using the first row:
\[ D = 1 \cdot \begin{vmatrix} c & a \\ b & c \end{vmatrix} - 1 \cdot \begin{vmatrix} a & b \\ b & c \end{vmatrix} + 1 \cdot \begin{vmatrix} a & b \\ c & a \end{vmatrix} \]
This simplifies to:
\[ D = (c \cdot c - a \cdot b) - (a \cdot c - b \cdot b) + (a \cdot a - b \cdot c) \]
\[ = c^2 - ab - ac + b^2 + a^2 - bc \]

3. **Setting the Determinant to Zero**: Since the determinant is zero, we have:
\[ c^2 - ab - ac + b^2 + a^2 - bc = 0 \]
Rearranging gives:
\[ a^2 + b^2 + c^2 - ab - ac - bc = 0 \]

4. **Recognizing the Condition**: The equation \( a^2 + b^2 + c^2 - ab - ac - bc = 0 \) indicates that the triangle is equilateral. This is because the equality holds if and only if \( a = b = c \).

5. **Finding Sine Values**: In an equilateral triangle, all angles \( A, B, C \) are equal to \( 60^\circ \). Therefore:
\[ \sin A = \sin B = \sin C = \sin 60^\circ = \frac{\sqrt{3}}{2} \]

6. **Calculating \( \sin^2 A + \sin^2 B + \sin^2 C \)**:
\[ \sin^2 A + \sin^2 B + \sin^2 C = 3 \cdot \left(\sin 60^\circ\right)^2 = 3 \cdot \left(\frac{\sqrt{3}}{2}\right)^2 \]
\[ = 3 \cdot \frac{3}{4} = \frac{9}{4} \]

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