(b) $\Delta \equiv \left| {\,\begin{array}{*{20}{c}}a&b&{a\alpha + b}\\b&c&{b\alpha + c}\\{a\alpha + b}&{b\alpha + c}&0\end{array}\,} \right|$

= $\left| {\,\begin{array}{*{20}{c}}a&b&{a\alpha + b}\\b&c&{b\alpha + c}\\0&0&{ – (a{\alpha ^2} + 2b\alpha + c)}\end{array}\,} \right|$, by ${R_3} \to {R_3} – \alpha {R_1} – {R_2}$

= $a\,\{ – c(a{\alpha ^2} + 2b\alpha + c) – 0\} – b\{ – b(a{\alpha ^2} + 2b\alpha + c) – 0\} $

by expanding along ${C_1}$

$ = ({b^2} – ac)\,(a{\alpha ^2} + 2b\alpha + c)$

Thus, $\Delta = 0$, if either ${b^2} – ac = 0$ or $a{\alpha ^2} + 2b\alpha + c = 0$

i.e., $a,b,c$ in $G.P.$ or $a{\alpha ^2} + 2b\alpha + c = 0$.

Trick: Put $\alpha = 0$, then the determinant

$\left| {\,\begin{array}{*{20}{c}}a&b&b\\b&c&c\\b&c&0\end{array}\,} \right|\, = \,\left| {\,\begin{array}{*{20}{c}}a&b&0\\b&c&0\\b&c&{ – c}\end{array}\,} \right|\, = \, – c(ac – {b^2}) = 0$.

Hence the result.