If the determinant $\begin{vmatrix}a&b&2a\, α + 3b\\b&c&2b\,α + 3c\\2a\, α + 3b&2b\,α + 3c&0\end{vmatrix}=0$, then
Answer & explanation
Correct answer: option 2
We have,
$\begin{vmatrix}a&b&2a\, α + 3b\\b&c&2b\,α + 3c\\2a\, α + 3b&2b\,α + 3c&0\end{vmatrix}=0$
Applying $C_3→C_3 -2α C_1 - 3C_2$, we get
$\begin{vmatrix}a&b&0\\b&c&0\\2a\, α + 3b&2b\,α + 3c&-2a(2a\, α + 3b)-3(2b\,α + 3c)\end{vmatrix}=0$
$⇒-(ac-b^2) \{2α (2a\, α + 3b) + 3 (2b\, α + 3c)\} = 0$
$⇒(b^2-ac)\{4a\, α^2 + 12b\, α + 9c\} = 0$
$⇒b^2-ac=0, 4a\, α^2 + 12b\, α + 9c=0$
⇒ a, b, c are in G.P. and $x = α$ is a root of $4ax^2 + 12bx+9c = 0$.