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(C) The given system of equations can be written as:$(a-1)x + ay + az = 0$$bx + (b-1)y + bz = 0$$cx + cy + (c-1)z = 0$Since $xyz 
eq 0$,the system ha

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(C) The given system of equations can be written as:$(a-1)x + ay + az = 0$$bx + (b-1)y + bz = 0$$cx + cy + (c-1)z = 0$Since $xyz 
eq 0$,the system has a non-trivial solution. Therefore,the determinant of the coefficient matrix must be zero:$\begin{vmatrix} a-1 & a & a \ b & b-1 & b \ c & c & c-1 \end{vmatrix} = 0$Applying the column operation $C_1 ightarrow C_1 + C_2 + C_3$:$\begin{vmatrix} a+b+c-1 & a & a \ a+b+c-1 & b-1 & b \ a+b+c-1 & c & c-1 \end{vmatrix} = 0$Taking $(a+b+c-1)$ as a common factor:$(a+b+c-1) \begin{vmatrix} 1 & a & a \ 1 & b-1 & b \ 1 & c & c-1 \end{vmatrix} = 0$Performing row operations $R_2 ightarrow R_2 - R_1$ and $R_3 ightarrow R_3 - R_1$:$(a+b+c-1) \begin{vmatrix} 1 & a & a \ 0 & -1 & 0 \ 0 & c-a & c-a-1 \end{vmatrix} = 0$Expanding the determinant along the first column:$(a+b+c-1) [1 	imes ((-1)(c-a-1) - 0)] = 0$$(a+b+c-1) (a-c+1) = 0$Since $x, y, z$ are non-zero,we must have $a+b+c-1 = 0$,which implies $a+b+c = 1$.

Given the system of equations $a(x + y + z) = x$,$b(x + y + z) = y$,$c(x + y + z) = z$ where $a, b, c$ are non-zero real numbers. If the real numbers $x, y, z$ are such that $xyz \neq 0$,then $(a + b + c)$ is equal to-

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