If the determinant

Solution:

(b) $\alpha$ is a root of $4 a x^{2}+12 b x+9 c=0$ or $a, b, c$ are in G.P.

Let $\Delta=\left|\begin{array}{lllllll}a & b & 2 a \alpha+3 b & b & c & 2 b \alpha+3 c 2 a \alpha+3 b & 2 b \alpha+3 c & 0\end{array}\right|$

$=\left|\begin{array}{lllllll}a-b & b & 2 a \alpha+3 b b-c & c & 2 b \alpha+3 c 2 a \alpha+3 b-2 b \alpha-3 c & 2 b \alpha+3 c\end{array}\right|$

[Applying $C_{1} \rightarrow C_{1}-C_{2}$ ]

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

$=\mid a-b \quad b \quad 2 a \alpha+3 b$

$b-c \quad c \quad 2 b \alpha+3 c$

$\begin{array}{lll}0 & 0 & -2 \alpha(2 a \alpha+3 b)-3(2 b \alpha+3 c)\end{array} \quad\left[\right.$ Applying $\left.R_{3} \rightarrow R_{3}-2 \alpha, R_{1}-3 R_{2}\right]$

$=-2 \alpha(2 a \alpha+3 b)-3(2 b \alpha+3 c) \mid a-b \quad b$

$b-c \quad c \mid \quad\left[\right.$ Expanding along $\left.R_{3}\right]$

$=-\left(4 a \alpha^{2}+12 b \alpha+9 c\right)\left(a c-b^{2}\right)$

But $\Delta=0$  [Given]

$\Rightarrow-\left(4 a \alpha^{2}+12 b \alpha+9 c\right)\left(a c-b^{2}\right)=0$

$\Rightarrow\left(4 a \alpha^{2}+12 b \alpha+9 c\right)=0$

or $\left(a c-b^{2}\right)=0$

$\Rightarrow \alpha$ is a root of $4 a x^{2}+12 b x+9 c=0$

or $a c=b^{2}$, i.e. $a, b, c$ are in $G . P$.