A non trivial solution of the system of equations $x+\lambda y+2z=0,2x+\lambda z=0,2\lambda x-2y+3z=0$ is given by $x:y:z=$

$\left|\begin{array}{l}1\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\lambda \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}2\\ 2 \hspace{0.25em} \hspace{0.25em}\hspace{0.25em}0\hspace{0.25em} \hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\lambda \\ 2\lambda \hspace{0.25em}-2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}3\end{array}\right|=0$

$\begin{array}{l}\Rightarrow 1\left(2\lambda \right)-\lambda (6-2\lambda )+2(-4)=0\Rightarrow {\lambda }^3-2\lambda -4=0\\ \Rightarrow (\lambda -2)\left({\lambda }^2+2\lambda +2\right)=0\end{array}$

Taking $\lambda =2$ the system becomes  $x+2y+2z=0$ and $2x+0\cdot y+2z=0$$,4x-2y+3z=0$

$\therefore \frac{x}{4}=\frac{y}{2}=\frac{z}{-4}\Rightarrow x:y:z=2:1:-2$