Let S be the set of all $\lambda \in R $ for which the system of linear equations

$2x-y + 2z=2$

$x-2y + \lambda z= - 4$

$x+ \lambda y + z = 4 $

has no solution, than the set S

contains exactly two elements

The correct answer is option (2) : contains exactly two elements

The given system of equations will have no solution, if D= 0 and at least one of $D_1, D_2, D_3 $ is non zero.

Now, $D=\begin{vmatrix}2& -1 & 2\\1 & -2 & \lambda\\1 & \lambda & 1\end{vmatrix}=-(\lambda -1) (2 \lambda +1)$

$D_1= \begin{vmatrix}2& -1 & 2\\-4 & -2 & \lambda\\4 & \lambda & 1\end{vmatrix}=-2 (\lambda^2 + 6\lambda - 4)$

$D_2=\begin{vmatrix}2& 2 & 2\\1 & -4 & \lambda\\1 & 4& 1\end{vmatrix}=-6(\lambda -1 )$

and, $D_3= \begin{vmatrix}2& -1 & 2\\1 & -2 & -4 \\1 & \lambda & 4\end{vmatrix}= 2 (5\lambda - 2)$

We observe that $D=0$ when $\lambda =-\frac{1}{2}, 1 $ and $D_1≠0, D_3≠0$ for these values of $\lambda.$ Hence , $S= \begin{Bmatrix}-\frac{1}{2}, 1\end{Bmatrix}$.