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(C) The given system of equations is:$(a-1)x - y - z = 0$$-x + (b-1)y - z = 0$$-x - y + (c-1)z = 0$For a non-trivial solution,the determinant of the c

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$abc$

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(C) The given system of equations is:$(a-1)x - y - z = 0$$-x + (b-1)y - z = 0$$-x - y + (c-1)z = 0$For a non-trivial solution,the determinant of the coefficient matrix must be zero:$\begin{vmatrix} a-1 & -1 & -1 \ -1 & b-1 & -1 \ -1 & -1 & c-1 \end{vmatrix} = 0$Adding $R_2$ and $R_3$ to $R_1$:$\begin{vmatrix} a-3 & b-3 & c-3 \ -1 & b-1 & -1 \ -1 & -1 & c-1 \end{vmatrix} = 0$Alternatively,expanding the determinant:$(a-1)((b-1)(c-1) - 1) + 1(-(c-1) - 1) - 1(1 + (b-1)) = 0$$(a-1)(bc - b - c + 1 - 1) + (-c + 1 - 1) - (1 + b - 1) = 0$$(a-1)(bc - b - c) - c - b = 0$$abc - ab - ac - bc + b + c - c - b = 0$$abc - ab - ac - bc = 0$Therefore,$ab + bc + ca = abc$.

If $a, b, c$ are non-zero real numbers and if the equations $(a-1) x=y+z, (b-1) y=z+x, (c-1) z=x+y$ have a non-trivial solution,then $ab+bc+ca=$

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