Given

$2x-y+2z=2$

$x-2y+z=-4$

$x + y + λz=4$, then the value of such that the given system of equations has no solution, is

1

The given system of equations will have no solution, if $D = 0$ and at least one of $D_1, D_2, D_3$ is non-zero, where

$D=\begin{vmatrix}2&-1&2\\1&-2&1\\1&1&λ\end{vmatrix},D_1=\begin{vmatrix}2&-1&2\\-4&-2&1\\4&1&λ\end{vmatrix},D_2=\begin{vmatrix}2&2&2\\1&-4&1\\1&4&λ\end{vmatrix}$ and, $D=\begin{vmatrix}2&-1&2\\1&-2&-4\\1&1&4\end{vmatrix}$

Now,

$D=0$

$⇒\begin{vmatrix}2&-1&2\\1&-2&1\\1&1&λ\end{vmatrix}=0$

$⇒2(-2λ-1)+(λ-1) + 2(1+2)=0$

$⇒-3λ+3=0$

$⇒λ=1$

For this value of λ, we have

$D_1=\begin{vmatrix}2&-1&2\\-4&-2&1\\4&1&1\end{vmatrix}=6-8+8≠0$

Hence, the system has no solution for λ=1