To solve the given problem, we begin by understanding the determinant condition presented:
The determinant of the matrix is given as:
| 1 | a | b |
| 1 | c | a |
| 1 | b | c |
The condition \(\begin{vmatrix} 1 & a & b \\ 1 & c & a \\ 1 & b & c \end{vmatrix} = 0\)implies that the rows (or columns) of the matrix are linearly dependent. Expanding the determinant, we have:
\(\begin{vmatrix} 1 & a & b \\ 1 & c & a \\ 1 & b & c \end{vmatrix} = 1 \cdot (c^2 - a^2) - a \cdot (b^2 - a^2) + b \cdot (b \cdot a - c \cdot a).\)
On simplifying,
\(= c^2 - a^2 - a b^2 + a^3 + b^2 a - b a^2\)
\(= c^2 - a^2 + a^3 - b a^2.\)
Since the determinant is zero, it indicates a relationship between sides of the triangle such that:
\((a - b)(b - c)(c - a) = 0\)
This implies the triangle is isosceles, with either \(a = b\), \(b = c\), or \(c = a\).
Next, we need to evaluate \(\sin^2 A + \sin^2 B + \sin^2 C.\)
In an equilateral triangle, the angles are all \(\frac{\pi}{3}\), and using the identity for sine, we can calculate the value:
\(\sin^2 A = \sin^2 B = \sin^2 C = \sin^2 \left(\frac{\pi}{3}\right) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}.\)
Therefore,
\(\sin^2 A + \sin^2 B + \sin^2 C = 3 \times \frac{3}{4} = \frac{9}{4}.\)
Thus, the correct answer is \(\frac{9}{4}\).