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(D) The system of equations has no solution if the determinant of the coefficient matrix $D = 0$ and at least one of the Cramer's rule determinants $(

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$-2$

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(D) The system of equations has no solution if the determinant of the coefficient matrix $D = 0$ and at least one of the Cramer's rule determinants $(D_x, D_y, D_z)$ is non-zero.First, calculate the determinant $D$ of the coefficient matrix:$D = \begin{vmatrix} 2 & -1 & -1 \ 1 & -2 & 1 \ 1 & 1 & \lambda \end{vmatrix}$$D = 2(-2\lambda - 1) - (-1)(\lambda - 1) + (-1)(1 - (-2))$$D = -4\lambda - 2 + \lambda - 1 - 3$$D = -3\lambda - 6$For the system to have no solution or infinitely many solutions, we set $D = 0$:$-3\lambda - 6 = 0 \Rightarrow \lambda = -2$Now, check for $\lambda = -2$ using $D_x$:$D_x = \begin{vmatrix} 12 & -1 & -1 \ -4 & -2 & 1 \ 4 & 1 & -2 \end{vmatrix} = 12(4 - 1) - (-1)(8 - 4) + (-1)(-4 + 8)$$D_x = 12(3) + 1(4) - 1(4) = 36 
eq 0$Since $D = 0$ and $D_x 
eq 0$, the system has no solution for $\lambda = -2$.

The value of $\lambda$ for which the system of equations $2x - y - z = 12,$ $x - 2y + z = -4,$ and $x + y + \lambda z = 4$ has no solution is

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