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(A) For the system of homogeneous linear equations to have non-zero solutions,the determinant of the coefficient matrix must be zero.$\Delta = \begin{

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$3$

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(A) For the system of homogeneous linear equations to have non-zero solutions,the determinant of the coefficient matrix must be zero.$\Delta = \begin{vmatrix} \lambda-1 & 3 \lambda+1 & 2 \lambda \ \lambda-1 & 4 \lambda-2 & \lambda+3 \ 2 & 3 \lambda+1 & 3 \lambda-3 \end{vmatrix} = 0$Applying row operations $R_1 ightarrow R_1 - R_2$ and $R_2 ightarrow R_2 - R_3$:$\Delta = \begin{vmatrix} 0 & 3-\lambda & \lambda-3 \ \lambda-3 & \lambda-3 & -2(\lambda-3) \ 2 & 3 \lambda+1 & 3 \lambda-3 \end{vmatrix} = 0$Taking $(\lambda-3)$ common from $R_1$ and $R_2$:$(\lambda-3)^2 \begin{vmatrix} 0 & -1 & 1 \ 1 & 1 & -2 \ 2 & 3 \lambda+1 & 3 \lambda-3 \end{vmatrix} = 0$Expanding the determinant:$(\lambda-3)^2 [0 - (-1)(3 \lambda-3 + 4) + 1(3 \lambda+1 - 2)] = 0$$(\lambda-3)^2 [3 \lambda+1 + 3 \lambda-1] = 0$$(\lambda-3)^2 [6 \lambda] = 0$This gives $\lambda = 3$ or $\lambda = 0$.The distinct values of $\lambda$ are $0$ and $3$.The sum of these values is $0 + 3 = 3$.

The sum of distinct values of $\lambda$ for which the system of equations
$(\lambda-1) x+(3 \lambda+1) y+2 \lambda z=0$
$(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0$
$2 x+(3 \lambda+1) y+3(\lambda-1) z=0$
has non-zero solutions,is

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The sum of distinct values of $\lambda$ for which the system of equations$(\lambda-1) x+(3 \lambda+1) y+2 \lambda z=0$$(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0$$2 x+(3 \lambda+1) y+3(\lambda-1) z=0$has non-zero solutions,is