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

$x+\lambda y+z=4$

For no solution :

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

$\Rightarrow 2\left(-2-\lambda^{2}\right)+1(1-\lambda)+2(\lambda+2)=0$

$\Rightarrow-2 \lambda^{2}+\lambda+1=0$

$\Rightarrow \lambda=1,-\frac{1}{2}$

$D_{x}=\left|\begin{array}{ccc}2 & -1 & 2 \\ -4 & 2 & \lambda \\ 4 & \lambda & 1\end{array}\right|=2\left|\begin{array}{ccc}1 & -1 & 2 \\ -2 & -2 & \lambda \\ \lambda & \lambda & 1\end{array}\right|$

$=2(1+\lambda)$

whichis not equal to zero for

$\lambda=1,-\frac{1}{2}$