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(D) The given system of equations is homogeneous: $-x + y \cos C + z \cos B = 0$,$x \cos C - y + z \cos A = 0$,$x \cos B + y \cos A - z = 0$.The syste

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a non-$0$ solution if $	riangle ABC$ is an equilateral triangle and not for all triangles

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(D) The given system of equations is homogeneous: $-x + y \cos C + z \cos B = 0$,$x \cos C - y + z \cos A = 0$,$x \cos B + y \cos A - z = 0$.The system has a non-$0$ solution if and only if the determinant of the coefficient matrix is $0$.Let $D = \begin{vmatrix} -1 & \cos C & \cos B \ \cos C & -1 & \cos A \ \cos B & \cos A & -1 \end{vmatrix}$.Expanding the determinant: $D = -1(1 - \cos^2 A) - \cos C(-\cos C - \cos A \cos B) + \cos B(\cos A \cos C + \cos B)$.$D = -1 + \cos^2 A + \cos^2 C + \cos A \cos B \cos C + \cos A \cos B \cos C + \cos^2 B$.$D = \cos^2 A + \cos^2 B + \cos^2 C + 2 \cos A \cos B \cos C - 1$.For any triangle $ABC$,the identity $\cos^2 A + \cos^2 B + \cos^2 C + 2 \cos A \cos B \cos C = 1$ holds if and only if the triangle is degenerate or specific conditions are met. However,for a non-degenerate triangle,$D = 0$ if and only if the triangle is equilateral $(A=B=C=60^\circ)$.If $A=B=C=60^\circ$,then $\cos A = \cos B = \cos C = 1/2$. Substituting these values,$D = 3(1/4) + 2(1/8) - 1 = 3/4 + 1/4 - 1 = 0$.Thus,the system has a non-$0$ solution only when the triangle is equilateral.

If $A, B, C$ are the angles of a triangle,then the system of equations $-x + y \cos C + z \cos B = 0$,$x \cos C - y + z \cos A = 0$,$x \cos B + y \cos A - z = 0$ have

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