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(B) The given system of equations can be written as:$(a - 1)x - y - z = 0$$-x + (b - 1)y - z = 0$$-x - y + (c - 1)z = 0$For a non-trivial solution,the

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(B) The given system of equations can be written as:$(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$Applying row operations $R_2 	o R_2 - R_3$ and $R_3 	o R_3 - R_1$:$\begin{vmatrix} a - 1 & -1 & -1 \ 0 & b & -c \ -a & 0 & c \end{vmatrix} = 0$Expanding along the first row:$(a - 1)(bc - 0) + 1(0 - ac) - 1(0 + ab) = 0$$(a - 1)(bc) - ac - ab = 0$$abc - bc - ac - ab = 0$$abc = ab + bc + ca$Thus,$ab + bc + ca = abc$.

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

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