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

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(C) Since the system of equations has a non-trivial solution,the determinant of the coefficient matrix must be zero:$\begin{vmatrix} a & 1 & 1 \ 1 & b & 1 \ 1 & 1 & c \end{vmatrix} = 0$Applying row operations $R_2 	o R_2 - R_1$ and $R_3 	o R_3 - R_1$:$\begin{vmatrix} a & 1 & 1 \ 1-a & b-1 & 0 \ 1-a & 0 & c-1 \end{vmatrix} = 0$Expanding along the first row:$a(b-1)(c-1) - 1(1-a)(c-1) + 1(0 - (1-a)(b-1)) = 0$$a(b-1)(c-1) - (1-a)(c-1) - (1-a)(b-1) = 0$Divide the entire equation by $(1-a)(1-b)(1-c)$ (noting $a, b, c 
eq 1$):$\frac{a(b-1)(c-1)}{(1-a)(1-b)(1-c)} - \frac{(1-a)(c-1)}{(1-a)(1-b)(1-c)} - \frac{(1-a)(b-1)}{(1-a)(1-b)(1-c)} = 0$$\frac{-a}{(1-a)} - \frac{1}{(1-b)} - \frac{1}{(1-c)} = 0$$\frac{a}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 0$Since $\frac{a}{1-a} = \frac{a-1+1}{1-a} = -1 + \frac{1}{1-a}$,we substitute this into the equation:$-1 + \frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 0$$\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 $x + y + cz = 0$,where $a, b, c \neq 1$,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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