If the system of equation $x-2y+z=-4,x+y+\lambda z=4$ and $2x-y+2z=2$has no solution, then $\lambda$ lies in

$$\begin{gathered} \mathrm{x}-2\mathrm{y}+\mathrm{z}=-4 ....\left(1\right) \\ 2\mathrm{x}-\mathrm{y}+2\mathrm{z}=2 ....\left(2\right) \end{gathered}$$

$x+y+\lambda z=4 ....\left(3\right)$

$\left(1\right)\times \left(2\right)-\left(2\right)\Rightarrow -\mathrm{y}=-10$

$\therefore \lambda =16-\mathrm{z}$

$\therefore$ put in equation (3)

$16-z+10+\lambda z=4$

$22=z(1-\lambda )$