If the system of equations

The given system of equations $2 x-y-z=12, x-2 y+z=-4$ and $x+y+\lambda z=4$ has no solution.

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

$\Rightarrow 2(-2 \lambda-1)-(-1)(\lambda-1)-1(1+2)=0$

$\Rightarrow-4 \lambda-2+\lambda-1-3=0$

$\Rightarrow-3 \lambda-6=0$

$\Rightarrow-3 \lambda=6$

$\Rightarrow \lambda=-2$

Thus, the value of $\lambda=-2$.

If the system of equations $2 x-y-z=12, x-2 y+z=-4, x+y+\lambda z=4$ has no solution, then $\lambda=$ $-2$