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(B) The given determinant is $\Delta = \begin{vmatrix} a & b & c \ b & c & a \ c & a & b \end{vmatrix} = 0$. Expanding the determinant: $a(cb - a^2)

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$\frac{9}{4}$

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(B) The given determinant is $\Delta = \begin{vmatrix} a & b & c \ b & c & a \ c & a & b \end{vmatrix} = 0$. Expanding the determinant: $a(cb - a^2) - b(b^2 - ac) + c(ba - c^2) = 0$. This simplifies to $abc - a^3 - b^3 + abc + abc - c^3 = 0$,which is $3abc - (a^3 + b^3 + c^3) = 0$. Rearranging gives $a^3 + b^3 + c^3 - 3abc = 0$. Using the identity $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - bc - ca) = 0$. Since $a, b, c$ are sides of a triangle,$a+b+c 
eq 0$. Thus,$a^2 + b^2 + c^2 - ab - bc - ca = 0$,which implies $\frac{1}{2}((a-b)^2 + (b-c)^2 + (c-a)^2) = 0$. This holds only if $a = b = c$. Since $a = b = c$,the triangle is equilateral,so $A = B = C = 60^\circ$. Therefore,$\sin^2 A + \sin^2 B + \sin^2 C = \sin^2 60^\circ + \sin^2 60^\circ + \sin^2 60^\circ = (\frac{\sqrt{3}}{2})^2 + (\frac{\sqrt{3}}{2})^2 + (\frac{\sqrt{3}}{2})^2 = \frac{3}{4} + \frac{3}{4} + \frac{3}{4} = \frac{9}{4}$.

If $a, b, c$ are sides of $\triangle ABC$ and $\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0$,then $\sin^2 A + \sin^2 B + \sin^2 C = $ . . . . . . .

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