Explanation

(b)
Since , $\left| \begin{matrix}
a & b & a\alpha + b \\
b & c & b\alpha + c \\
a\alpha + b & b\alpha + c & 0
\end{matrix} \right| = 0$
Applying $R_{3} \rightarrow R_{3} - (\alpha R_{1} + R_{2})$
$\Rightarrow \left| \begin{matrix}
a & b & a\alpha + b \\
b & c & b\alpha + c \\
0 & 0 & - a\alpha^{2} - 2b\alpha - c
\end{matrix} \right| = 0$
$\Rightarrow - \left( a\alpha^{2} + 2b\alpha + c \right)\left( ac + b^{2} \right) = 0$
$\Rightarrow b^{2} = ac$
Hence, $a,b$, and $c$ are in GP.