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