Correct option is (2) \((-3,0)\)
\(\left|\begin{array}{ccc}
1 & \frac{3}{2} & \alpha+\frac{3}{2} \\
1 & \frac{1}{3} & \alpha+\frac{1}{3} \\
2 \alpha+3 & 3 \alpha+1 & 0
\end{array}\right|=0\)
\(\Rightarrow(2 \alpha+3)\left\{\frac{7 \alpha}{6}\right\}-(3 \alpha+1)\left\{\frac{-7}{6}\right\}=0\)
\(\Rightarrow(2 \alpha+3) \cdot \frac{7 \alpha}{6}+(3 \alpha+1) \cdot \frac{7}{6}=0\)
\(\Rightarrow 2 \alpha^{2}+3 \alpha+3 \alpha+1=0\)
\( \Rightarrow 2 \alpha^{2}+6 \alpha+1=0\)
\(\Rightarrow \alpha=\frac{-3+\sqrt{7}}{2}, \frac{-3-\sqrt{7}}{2}\)