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(C) For the system of equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero:$\begin{vmatrix} a & 1 & 1 \ 1

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$1$

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(C) For the system of equations to have a non-trivial solution,the determinant of the coefficient matrix must be zero:$\begin{vmatrix} a & 1 & 1 \ 1 & b & 1 \ 1 & 1 & c \end{vmatrix} = 0$Expanding the determinant:$a(bc - 1) - 1(c - 1) + 1(1 - b) = 0$$abc - a - c + 1 + 1 - b = 0$$abc - (a + b + c) + 2 = 0$Alternatively,perform row operations:$R_1 	o R_1 - R_2$ and $R_2 	o R_2 - R_3$$\begin{vmatrix} a-1 & 1-b & 0 \ 0 & b-1 & 1-c \ 1 & 1 & c \end{vmatrix} = 0$$(a-1)[(b-1)c - (1-c)] - (1-b)[0 - (1-c)] = 0$$(a-1)(bc - b - 1 + c) + (1-b)(1-c) = 0$Dividing by $(1-a)(1-b)(1-c)$ (since $a, b, c 
eq 1$):$\frac{-(a-1)(bc - b - 1 + c)}{(1-a)(1-b)(1-c)} + \frac{(1-b)(1-c)}{(1-a)(1-b)(1-c)} = 0$$\frac{bc - b - 1 + c}{(1-b)(1-c)} + \frac{1}{1-a} = 0$$\frac{bc - b - 1 + c + 1 - b - c + bc}{(1-b)(1-c)} = 0$ (This simplifies to the identity $\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$).

If the system of equations $ax + y + z = 0$,$x + by + z = 0$,and $x + y + cz = 0$ $(a, b, c \neq 1)$ has a non-trivial solution,then the value of $\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$ is:

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