For a square linear system to have no solution, the coefficient determinant $\Delta =0$ while at least one of ${\Delta }_x,{\Delta }_y,{\Delta }_z\ne 0$.

$\Delta =\left|\begin{array}{c}314\\ 2\alpha -1\\ 121\end{array}\right|=3(\alpha +2)+1(-1-(-2))\cdot (-1)+4(4-\alpha )$.

Carefully: expand along row 1: $3[\alpha (1)-(-1)(2)]-1[2(1)-(-1)(1)]+4[2(2)-\alpha (1)]=3(\alpha +2)-1(3)+4(4-\alpha )=3\alpha +6-3+16-4\alpha =-\alpha +19$.

$\Delta =0\Rightarrow \alpha =19$.

Check ${\Delta }_x\ne 0$ at $\alpha =19$: replace first column with RHS $(3,-3,4{)}^T$; computation gives a non-zero value, confirming ‘no solution’.