b
(b) $\left| {\,\begin{array}{*{20}{c}}a&b&{a\alpha - b}\\b&c&{b\alpha - c}\\2&1&0\end{array}\,} \right| = 0$
==> $a[ - (b\alpha - c)] - b[ - 2(b\alpha - c)] + [a\alpha - b)(b - 2c)] = 0$
==>$ - ab\alpha + ac + 2{b^2}\alpha - 2bc + ab\alpha - 2ac\alpha - {b^2} + 2bc = 0$
==> $ac + 2{b^2}\alpha - 2ac\alpha - {b^2} = 0$
==> $(ac - {b^2}) - 2\alpha (ac - {b^2}) = 0$
==> $ac - {b^2} = 0$or $1 - 2\alpha = 0$
$ \Rightarrow $ ${b^2} = ac$ or $\alpha = \frac{1}{2}$
(As given in question)
So, ${b^2} = ac$ i.e, $a,b,c$ are in $G.P.$