If a, b, c are the sides of a triangle ABC an | vedclass

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(C) Given the determinant equation: $\left|\begin{array}{lll}b & 1 & a \ a & 1 & c \ c & 1 & b\end{array}ight| = 0$.Expanding the determinant alon

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$3$

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(C) Given the determinant equation: $\left|\begin{array}{lll}b & 1 & a \ a & 1 & c \ c & 1 & b\end{array}ight| = 0$.Expanding the determinant along the first row $(R_1)$:$b(b - c) - 1(ab - c^2) + a(a - c) = 0$$b^2 - bc - ab + c^2 + a^2 - ac = 0$$a^2 + b^2 + c^2 - (ab + bc + ca) = 0 \quad \dots(i)$Multiplying by $2$ on both sides:$2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca = 0$$(a - b)^2 + (b - c)^2 + (c - a)^2 = 0$Since the sum of squares is zero,each term must be zero: $a = b = c$.Thus,$	riangle ABC$ is an equilateral triangle,so $A = B = C = 60^{\circ}$.Now,calculate $2(\cos A + \cos B + \cos C) = 2(\cos 60^{\circ} + \cos 60^{\circ} + \cos 60^{\circ})$$= 2(3 	imes \frac{1}{2}) = 3$.

If $a, b, c$ are the sides of a triangle $ABC$ and $2(\cos A + \cos B + \cos C) = \left|\begin{array}{lll}b & 1 & a \\ a & 1 & c \\ c & 1 & b\end{array}\right| = 0$,then find the value of the expression.

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