The set of all values of $\lambda$ for which the system of linear equations $2x_1-2x_2+x_3=\lambda x_1,2x_1-3x_2+2x_3=\lambda x_2$ and $-x_1+2x_2=\lambda x_3$ has a non - tirvial solution.

The correct option is C

contains two elements

Given system of linear equations
$2x_1-2x_2+x_3=\lambda x_1$
$\Rightarrow (2-\lambda )x_1-2x_2+x_3=0$ …(i)
$2x_1-3x_2+2x_3=\lambda x_2$
$\Rightarrow 2x_1-(3+\lambda )x_2+2x_3=0$ …(ii)
$-x_1+2x_2=\lambda x_3$
$\Rightarrow -x_1+2x_2-\lambda x_3=0$ …(iii)
Since, the system has non - trival solution.
$\therefore \left|\begin{array}{ccc}2-\lambda & -2& 1\\ 2& -(3+\lambda )& 2\\ -1& 2& -\lambda \end{array}\right|=0$
$\Rightarrow (2-\lambda )(3\lambda +{\lambda }^2-4)+2(-2\lambda +2)+1(4-3)-\lambda =0$
$\Rightarrow (2-\lambda )({\lambda }^2+3\lambda -4)+4(1-\lambda )+(1-\lambda )=0$
$\Rightarrow (2-\lambda )(\lambda +4)(\lambda -1)+5(1-\lambda )=0$
$\Rightarrow (\lambda -1)\left[\right(2-\lambda \left)\right(\lambda +4)-5]=0$
$\Rightarrow (\lambda -1)({\lambda }^2+2\lambda -3)=0$
$\Rightarrow (\lambda -1)\left[\right(\lambda -1\left)\right(\lambda +3\left)\right]=0$
$\Rightarrow (\lambda -1{)}^2(\lambda +3)=0$
$\Rightarrow \lambda =1,1,-3$
Hence, $\lambda$ contains two elements.