The system of linear equations $x-y+z=1, x+y-z=3, x-4y+4z=\alpha$has

given$x-y+z=1\_\_\_\_\_\_\left(1\right)$

$x+y-z=3\_\_\_\_\_\_\left(2\right)$

$x-4y+4z=\alpha \_\_\_\_\_\_\left(3\right)$

From eqns 1 & 2     $x=2\Rightarrow$y-z=1 & $-4y+4z=\alpha -2$

$4y-4z+\left(-4y+4z\right)=4+\left(\alpha -2\right)\Rightarrow \alpha +2=0\Rightarrow \alpha =-2$

The system of has no solution for$\alpha \ne -2$

   If $\alpha =-2,$ then the system is consistent and has infinite number of solutions. (since y-z=1 has infinite number of solutions)