The values of $\lambda $ and $\mu $ for which the system of linear equations

$x+y + z= 2 $

$x+ 2y + 3z= 5 $

$x+ 3y + \lambda z = \mu $

has infinitely many solutions are respectively

5 and 8

The correct answer is option (1) : 5 and 8

For the given system of equations, we have

$D=\begin{vmatrix}1 & 1 & 1\\1 & 2 & 3\\1 & 3 & λ\end{vmatrix}= λ-5, D_3= \begin{vmatrix}1 & 1 & 2\\1 & 2 & 5\\1 & 3 & \mu \end{vmatrix}=\mu - 8 $,

$D_1= \begin{vmatrix}2 & 1 & 1\\5 & 2 & 3\\\mu  & 3 & λ\end{vmatrix}= - λ+ \mu - 3, D_2= \begin{vmatrix}1& 2 & 1\\1 & 5 & 3\\1 & \mu & λ\end{vmatrix}=3λ-2\mu + 1$

If the system of equations has infinitely many solutions,

then we find that $D_1=D_2=D_3=D=0$ for $\lambda = 5 $ and $\mu = 8.$

Hence, the given system of equations has infinitely many solutions for $\lambda = 5 $ and $\mu = 8.$