In a $\triangle ABC$, if $$\begin{vmatrix} 1 & a & b \\ 1 & c & a \\ 1 & b & c \end{vmatrix} = 0$$, then $\sin^2 A + \sin^2 B + \sin^2 C = $_____.
Step-by-Step Solution
Key Concept: Recognize that the determinant condition ∣1 a b; 1 c a; 1 b c∣ = 0 leads to a² + b² + c² - ab - bc - ca = 0, which factors as (a-b)² + (b-c)² + (c-a)² = 0, forcing a = b = c (equilateral triangle). Then apply the sine rule with Law of Cosines to find sin²A + sin²B + sin²C = 3sin²60° = 3(√3/2)² = 9/4.
From the determinant condition $\begin{vmatrix}1 & a & b\\c & a & 0\\1 & b & c\end{vmatrix} = 0$, expanding gives $a^2 + b^2 + c^2 - ab - bc - ca = 0$, which factors as $(a-b)^2 + (b-c)^2 + (c-a)^2 = 0$. This implies $a = b = c$, so triangle $ABC$ is equilateral with all angles $60°$. Therefore $\sin^2 A + \sin^2 B + \sin^2 C = 3\sin^2 60° = \frac{9}{4}$.
Correct Answer: 2.25