If the determinant

$\begin{vmatrix}a & b & 2a \alpha  + 3b\\b & c & 2b \alpha + 3c\\2a \alpha + 3b & 2b\alpha + 3c & 0\end{vmatrix}=0,$ then

$\alpha $ is a root of $4ax^2 + 12bx + 9c =0, $ or, $ a, b, c are in G.P.

The correct answer is option (2) : $\alpha $ is a root of $4ax^2 + 12bx + 9c =0, $ or, $ a, b, c are in G.P.

We have,

$\begin{vmatrix}a & b & 2a \alpha  + 3b\\b & c & 2b \alpha + 3c\\2a \alpha + 3b & 2b\alpha + 3c & 0\end{vmatrix}=0$

Applying $C_3→C_3-2\alpha c_1= 3C_2, $ we get

$\begin{vmatrix}a & b & 0\\b & c & 0\\2a \alpha + 3b & 2b\alpha + 3c &-2\alpha (2a \alpha + 3b) -3 (2b \alpha + 3c)\end{vmatrix}=0$

$⇒-(ac-b^2) \begin{Bmatrix} 2\alpha (2a \alpha + 3b) + 3 (2b \alpha + 3c )\end{Bmatrix}=0$

$⇒(b^2 - ac)  \begin{Bmatrix}4a \alpha^2 + 12 b \alpha + 9c \end{Bmatrix}= 0 $

$⇒b^2 - ac= 0, 4a \alpha^2 + 12b \alpha + 9 c= 0 $

⇒ a, b, c are in G.P and $ x= \alpha $ is a root of $4ax^2 + 12bx + 9c = 0 .$