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-trivial solution

Given system of linear equations 

$\begin{array}{l}\Rightarrow (2-\lambda )x_1-2x_2+x_3=0----\left(1\right)\\ \begin{array}{rl} & 2x_1-3x_2+2x_3=\lambda x_2----\left(2\right)\\ & \Rightarrow 2x_1-(3+\lambda )x_2+2x_3=0----\left(3\right)\\ & -x_1+2x_2=\lambda x_3\\ & \Rightarrow -x_1+2x_2-\lambda x_3=0\end{array}\end{array}$

Since , the system has non-trivial solution

$\begin{array}{l}\therefore \left[\begin{array}{ccc}2-\lambda & -2& 1\\ 2& -(3+\lambda )& 2\\ -1& 2& \lambda \end{array}\right]=0\\ \left.\Rightarrow (2-\lambda )\left(3\lambda +{\lambda }^2-4\right)+2(-2\lambda +2)+1(4-3)-\lambda \right)=0\\ \Rightarrow (2-\lambda )\left({\lambda }^2+3\lambda -4\right)+4(1-\lambda )+(1-\lambda )=0\\ \Rightarrow (2-\lambda )(\lambda +4)(\lambda -1)+5(1-\lambda )=0\\ \Rightarrow (\lambda -1)\left[\right(2-\lambda \left)\right(\lambda +4)-5]=0\\ \Rightarrow (\lambda -1)\left({\lambda }^2+2\lambda -3\right)=0\\ \Rightarrow (\lambda -1)\left[\right(\lambda -1\left)\right(\lambda +8\left)\right]=0\\ \Rightarrow (\lambda -1{)}^2(\lambda +3)=0\\ \Rightarrow \lambda =1,1,-3\end{array}$

Hence , $\lambda$ contains two elements.