The value of $\lambda$ such that the system $x-2y+z=-4,$$2x-y+2z=2,x+y+\lambda z=4$ has no solution is  

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

$=1(-\lambda -2)-2(2\lambda -2)+1(2+1)$
$\Rightarrow \mathrm{\Delta }=-5(\lambda -1)$

${\mathrm{\Delta }}_x=\left|\begin{array}{ccc}-4& -2& 1\\ 2& -1& 2\\ 4& 1& \lambda \end{array}\right|$

${\mathrm{\Delta }}_x=-4(-\lambda -2)+2(-2\lambda +8)+1(2+4)=30$

${\mathrm{\Delta }}_y=\left|\begin{array}{ccc}1& -4& 1\\ 2& 2& 2\\ 1& 4& \lambda \end{array}\right|$

$=1(2\lambda -8)+4(2-2\lambda )+1(8-2)=-6\lambda +6$

${\mathrm{\Delta }}_z=\left|\begin{array}{ccc}1& -2& -4\\ 2& -1& 2\\ 1& 1& 4\end{array}\right|$

$=1(-4-2)+2(2-8)-4(2+1)=-30$

For no solution, we must have $∆= 0$and atleast one

${\mathrm{\Delta }}_x,{\mathrm{\Delta }}_y$and ${\mathrm{\Delta }}_z$non-zero 

Clearly for $\lambda =1,\mathrm{\Delta }=0$ and ${\mathrm{\Delta }}_x={\mathrm{\Delta }}_z\ne 0$