The values of alpha,for which left|begin{arra | vedclass

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(B) Given the determinant equation: $\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \ 2 \alpha+

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$(-3, 0)$

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(B) Given the determinant equation: $\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}ight|=0$ Expanding along the third row: $(2\alpha+3) \left( \frac{3}{2}(\alpha+\frac{1}{3}) - \frac{1}{3}(\alpha+\frac{3}{2}) ight) - (3\alpha+1) \left( 1(\alpha+\frac{1}{3}) - 1(\alpha+\frac{3}{2}) ight) = 0$ Simplify the terms inside the brackets: $(2\alpha+3) \left( \frac{3\alpha}{2} + \frac{1}{2} - \frac{\alpha}{3} - \frac{1}{2} ight) - (3\alpha+1) \left( \alpha + \frac{1}{3} - \alpha - \frac{3}{2} ight) = 0$ $(2\alpha+3) \left( \frac{9\alpha - 2\alpha}{6} ight) - (3\alpha+1) \left( \frac{2 - 9}{6} ight) = 0$ $(2\alpha+3) \left( \frac{7\alpha}{6} ight) - (3\alpha+1) \left( -\frac{7}{6} ight) = 0$ Divide by $\frac{7}{6}$: $(2\alpha+3)(\alpha) + (3\alpha+1) = 0$ $2\alpha^2 + 3\alpha + 3\alpha + 1 = 0$ $2\alpha^2 + 6\alpha + 1 = 0$ Using the quadratic formula $\alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $\alpha = \frac{-6 \pm \sqrt{36 - 8}}{4} = \frac{-6 \pm \sqrt{28}}{4} = \frac{-6 \pm 2\sqrt{7}}{4} = \frac{-3 \pm \sqrt{7}}{2}$ Since $\sqrt{7} \approx 2.645$,the values are $\alpha_1 = \frac{-3 + 2.645}{2} \approx -0.1775$ and $\alpha_2 = \frac{-3 - 2.645}{2} \approx -2.8225$. Both values lie in the interval $(-3, 0)$. Thus,option $B$ is correct.

The values of $\alpha$,for which $\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$,lie in the interval

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