The set of all values of $\lambda$ for which the system of linear equation :
$2x_1-2x_2+x_3=\lambda x_1,2x_1-3x_2+2x_3=\lambda x_2-x_1+2x_2=\lambda x_3$ has a non-trivial solution

The correct option is A contains two elements
$(2-\lambda )x_1-2x_2+x_3=0$

$2x_1-(3+\lambda )x^2+2x_3=0$

$-x_1+2x_2-\lambda x_3=0$

Non-Trivial solution $△=0$

$\left|\begin{array}{ccc}2-\lambda & -2& 1\\ 2& -3-\lambda & 2\\ -1& 2& -\lambda \end{array}\right|=0$

$\Rightarrow (2-\lambda )\{3\lambda +{\lambda }^2-4\}+2[-2\lambda +2\}+(4-3-\lambda )=0$

$\Rightarrow (6\lambda +2{\lambda }^2-8-3{\lambda }^2-{\lambda }^3+4\lambda )-4\lambda +4+1-\lambda =0$

$\Rightarrow -{\lambda }^3-{\lambda }^2-5\lambda +3=0$

$x^3-{\lambda }^2+2{\lambda }^2-2\lambda -3\lambda +3=0$

${\lambda }^2(\lambda -1)+2\lambda (\lambda -1)-3(\lambda -1)=0$

$(\lambda -1)=({\lambda }^2+2\lambda -3)=0$

$(\lambda -1)(\lambda +3)(\lambda -1)=0$

$\lambda =1,1,-3$

Therefore, thus are only Two elements.