Consider the following system of linear equations:
$x+2 y+z=1$
$2 x+y+z=\alpha$
$4 \mathrm{x}+5 \mathrm{y}+3 \mathrm{z}=\alpha^{2}$. Then the system has
(a) infinitely many solutions when $\alpha=-1$ or 2
(b) infinitely many solutions when $\alpha=1$ or $-2$
(c) no solution when $\alpha \in \mathrm{R}-\{-1,2\}$
(d) no solution when $\alpha \in \mathrm{R}-\{1,-2\}$