Given $$\begin{vmatrix} 1 & \frac{3}{2} & \alpha + \frac{3}{2} \\ 1 & \frac{1}{3} & \alpha + \frac{1}{3} \\ 2\alpha + 3 & 3\alpha + 1 & 0 \end{vmatrix} = 0$$
First operation will be $$C_2$$ $$\Rightarrow$$ $$C_2-C_3$$
$$\therefore$$ $$\begin{vmatrix} 1 & \frac{3}{2}- \alpha - \frac{3}{2} & \alpha + \frac{3}{2} \\ 1 & \frac{1}{3}-\alpha - \frac{1}{3} & \alpha + \frac{1}{3} \\ 2\alpha + 3 & 3\alpha + 1 & 0 \end{vmatrix} = 0$$
$$\therefore$$ $$\begin{vmatrix} 1 & - \alpha & \alpha + \frac{3}{2} \\ 1 & -\alpha & \alpha + \frac{1}{3} \\ 2\alpha + 3 & 3\alpha + 1-0 & 0 \end{vmatrix} = 0$$
Now we will operate $$R_1$$ $$\Rightarrow$$ $$R_1-R_2$$
$$\therefore$$ $$\begin{vmatrix} 1-1 & \alpha- \alpha & \alpha + \frac{3}{2}-\alpha - \frac{1}{3} \\ 1 & -\alpha & \alpha + \frac{1}{3} \\ 2\alpha + 3 & 3\alpha + 1 & 0 \end{vmatrix} = 0$$
$$\therefore$$ $$\begin{vmatrix} 0 & 0 & \frac{7}{6} \\ 1 & -\alpha & \alpha + \frac{1}{3} \\ 2\alpha + 3 & 3\alpha + 1 & 0 \end{vmatrix} = 0$$
$$\therefore$$ $$\dfrac{7}{6}\left(\left(1\right)\left(3\alpha\ +1\right)-\left(-\alpha\ \right)\left(2\alpha\ +3\right)\right)=0$$
$$\therefore$$ $$2\alpha\ ^2+6\alpha\ +1=0$$
$$\therefore$$ $$\alpha\ =\dfrac{\left(-6\pm\ \sqrt{\ 28}\right)}{4}$$
$$\therefore$$ $$\alpha_1\ =-1.5+\dfrac{\sqrt{\ 7}}{2}$$ and $$\alpha_2\ =-1.5-\dfrac{\sqrt{\ 7}}{2}$$
$$\sqrt{\ 7}\simeq\ 2.645$$
$$\therefore$$ $$\alpha\ _1=-0.177$$ and $$\alpha\ _2=-2.8228$$
$$\therefore$$ $$\alpha\ \in\ \left(0,-3\right)$$
Hence, correct option is B.
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