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(D) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ must be $0$,and the determinants

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$5$ and $8$

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(D) For a system of linear equations to have infinitely many solutions,the determinant of the coefficient matrix $D$ must be $0$,and the determinants $D_1, D_2, D_3$ must also be $0$.First,we calculate $D = \begin{vmatrix} 1 & 1 & 1 \ 1 & 2 & 3 \ 1 & 3 & \lambda \end{vmatrix} = 0$.Expanding along the first row: $1(2\lambda - 9) - 1(\lambda - 3) + 1(3 - 2) = 0$.$2\lambda - 9 - \lambda + 3 + 1 = 0 \implies \lambda - 5 = 0 \implies \lambda = 5$.Next,we set $D_1 = \begin{vmatrix} 2 & 1 & 1 \ 5 & 2 & 3 \ \mu & 3 & 5 \end{vmatrix} = 0$.Expanding along the first row: $2(10 - 9) - 1(25 - 3\mu) + 1(15 - 2\mu) = 0$.$2(1) - 25 + 3\mu + 15 - 2\mu = 0$.$2 - 10 + \mu = 0 \implies \mu - 8 = 0 \implies \mu = 8$.Thus,the values are $\lambda = 5$ and $\mu = 8$.

The values of $\lambda$ and $\mu$ for which the system of linear equations $x+y+z=2$,$x+2y+3z=5$,and $x+3y+\lambda z=\mu$ has infinitely many solutions are,respectively:

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