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(C) For the system of linear equations to have no solution,the determinant of the coefficient matrix $\Delta$ must be $0$,and at least one of the Cram

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$24$

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(C) For the system of linear equations to have no solution,the determinant of the coefficient matrix $\Delta$ must be $0$,and at least one of the Cramer's determinants $\Delta_x, \Delta_y, \Delta_z$ must be non-zero.First,calculate $\Delta = \begin{vmatrix} 6 \lambda & -3 & 3 \ 2 & 6 \lambda & 4 \ 3 & 2 & 3 \lambda \end{vmatrix} = 0$.Expanding along the first row:$\Delta = 6 \lambda (18 \lambda^2 - 8) + 3 (6 \lambda - 12) + 3 (4 - 18 \lambda) = 0$$108 \lambda^3 - 48 \lambda + 18 \lambda - 36 + 12 - 54 \lambda = 0$$108 \lambda^3 - 84 \lambda - 24 = 0$Dividing by $12$: $9 \lambda^3 - 7 \lambda - 2 = 0$.By inspection,$\lambda = 1$ is a root. Using synthetic division,$( \lambda - 1 )( 9 \lambda^2 + 9 \lambda + 2 ) = 0$.$( \lambda - 1 )( 3 \lambda + 1 )( 3 \lambda + 2 ) = 0$.So,$\lambda \in \{ 1, -1/3, -2/3 \}$.For these values,we check $\Delta_1 = \begin{vmatrix} 4 \lambda^2 & -3 & 3 \ 1 & 6 \lambda & 4 \ \lambda & 2 & 3 \lambda \end{vmatrix} 
eq 0$.For $\lambda = 1$,$\Delta_1 = \begin{vmatrix} 4 & -3 & 3 \ 1 & 6 & 4 \ 1 & 2 & 3 \end{vmatrix} = 4(18-8) + 3(3-4) + 3(2-6) = 40 - 3 - 12 = 25 
eq 0$.For $\lambda = -1/3$,$\Delta_1 
eq 0$. For $\lambda = -2/3$,$\Delta_1 
eq 0$.Thus,$S = \{ 1, -1/3, -2/3 \}$.$12 \sum_{\lambda \in S} |\lambda| = 12 ( |1| + |-1/3| + |-2/3| ) = 12 ( 1 + 1/3 + 2/3 ) = 12 ( 2 ) = 24$.

Let $S$ be the set of values of $\lambda$,for which the system of equations
$6 \lambda x - 3 y + 3 z = 4 \lambda^2$
$2 x + 6 \lambda y + 4 z = 1$
$3 x + 2 y + 3 \lambda z = \lambda$
has no solution. Then $12 \sum_{\lambda \in S} |\lambda|$ is equal to $...........$.

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Let $S$ be the set of values of $\lambda$,for which the system of equations$6 \lambda x - 3 y + 3 z = 4 \lambda^2$$2 x + 6 \lambda y + 4 z = 1$$3 x + 2 y + 3 \lambda z = \lambda$has no solution. Then $12 \sum_{\lambda \in S} |\lambda|$ is equal to $...........$.