If (2alpha + 1, alpha^2 - 3alpha, frac{alpha | vedclass

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(D) Let $P = (\alpha, 2\alpha, 1)$ and $P' = (2\alpha + 1, \alpha^2 - 3\alpha, \frac{\alpha - 1}{2})$.The midpoint $M$ of $PP'$ is $(\frac{3\alpha+1}{

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Only $3$ and $\frac{1}{4}$

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(D) Let $P = (\alpha, 2\alpha, 1)$ and $P' = (2\alpha + 1, \alpha^2 - 3\alpha, \frac{\alpha - 1}{2})$.The midpoint $M$ of $PP'$ is $(\frac{3\alpha+1}{2}, \frac{\alpha^2-\alpha}{2}, \frac{\alpha+1}{4})$.Since $M$ lies on the line $\frac{x-2}{3} = \frac{y-1}{2} = \frac{z}{1}$, we have $\frac{\frac{3\alpha+1}{2} - 2}{3} = \frac{\frac{\alpha^2-\alpha}{2} - 1}{2} = \frac{\frac{\alpha+1}{4}}{1}$.Solving $\frac{3\alpha-3}{6} = \frac{\alpha+1}{4}$ gives $12\alpha - 12 = 6\alpha + 6$, so $6\alpha = 18$, which implies $\alpha = 3$.Also, the vector $\vec{PP'} = (\alpha+1, \alpha^2-5\alpha, \frac{\alpha-3}{2})$ must be perpendicular to the line direction $\vec{v} = (3, 2, 1)$.Thus, $3(\alpha+1) + 2(\alpha^2-5\alpha) + 1(\frac{\alpha-3}{2}) = 0$.Multiplying by $2$: $6\alpha + 6 + 4\alpha^2 - 20\alpha + \alpha - 3 = 0$, which simplifies to $4\alpha^2 - 13\alpha + 3 = 0$.Factoring gives $(4\alpha - 1)(\alpha - 3) = 0$, so $\alpha = 3$ or $\alpha = 1/4$.Both conditions are satisfied by $\alpha = 3$ and $\alpha = 1/4$.

If $(2\alpha + 1, \alpha^2 - 3\alpha, \frac{\alpha - 1}{2})$ is the image of $(\alpha, 2\alpha, 1)$ in the line $\frac{x-2}{3} = \frac{y-1}{2} = \frac{z}{1}$, then the possible value$(s)$ of $\alpha$ is (are)

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