Explanation

(d)
Applying $C_{3} \rightarrow C_{3} - {\alpha C}_{1} + C_{2}$ to the given determinant, we get
$\left| \begin{matrix}
a & b & 0 \\
b & c & 0 \\
2 & 1 & - 2\alpha + 1
\end{matrix} \right| = (1 - 2\alpha)(ac - b^{2})$
So, if the determinant is zero, we must have
$(1 - 2\alpha)\left( ac - b^{2} \right) = 0$
$\Rightarrow \ 1 - 2\alpha = 0$
or $\left( ac - b^{2} \right) = 0$
$\Rightarrow \ \alpha = \frac{1}{2}\ or\ ac = b^{2}$
Which means $a,\ b,\ c$ are in GP.