If (2α+1, α^2-3α, (α-1)/(2)) is the image of (α, 2α, 1) in the line...

Let the given point be $P = (\alpha, 2\alpha, 1) and its image be$P' = \left(2\alpha+1, \alpha^2-3\alpha, \dfrac{\alpha-1}{2}\right)$.<br><br>For$P'$to be the image of$P$in the given line$L: \dfrac{x-2}{3}=\dfrac{y-1}{2}=\dfrac{z}{1}$, two conditions must be satisfied:<br>1. The midpoint of$P$and$P'$must lie on the line$L$.<br>2. The line segment$PP'$must be perpendicular to the line$L$.<br><br>The midpoint$M$of$P$and$P'$is:<br>$M = \left( \dfrac{\alpha + 2\alpha + 1}{2}, \dfrac{2\alpha + \alpha^2 - 3\alpha}{2}, \dfrac{1 + \dfrac{\alpha-1}{2}}{2} \right) = \left( \dfrac{3\alpha + 1}{2}, \dfrac{\alpha^2 - \alpha}{2}, \dfrac{\alpha + 1}{4} \right)$<br><br>Since$M$lies on$L$, its coordinates must satisfy the equation of the line:<br>$\dfrac{\dfrac{3\alpha + 1}{2} - 2}{3} = \dfrac{\dfrac{\alpha^2 - \alpha}{2} - 1}{2} = \dfrac{\dfrac{\alpha + 1}{4}}{1}$<br><br>Simplifying the terms, we get:<br>$\dfrac{\alpha - 1}{2} = \dfrac{\alpha^2 - \alpha - 2}{4} = \dfrac{\alpha + 1}{4}$<br><br>Equating the first and third parts:<br>$\dfrac{\alpha - 1}{2} = \dfrac{\alpha + 1}{4}$<br><br>$2(\alpha - 1) = \alpha + 1 \Rightarrow 2\alpha - 2 = \alpha + 1 \Rightarrow \alpha = 3$<br><br>Substituting$\alpha = 3$into the middle part to verify:<br>$\dfrac{3^2 - 3 - 2}{4} = \dfrac{4}{4} = 1$<br>The first and third parts also equal$1$, so$\alpha = 3$satisfies the midpoint condition.<br><br>Now, let us check the perpendicularity condition. The direction ratios of$\overrightarrow{PP'}$are:<br>\left(2\alpha+1 - \alpha, \alpha^2-3\alpha - 2\alpha, \dfrac{\alpha-1}{2} - 1\right) = \left(\alpha+1, \alpha^2-5\alpha, \dfrac{\alpha-3}{2}\right)$

The direction ratios of the line $L$ are $(3, 2, 1)$. The dot product must be zero:
$3(\alpha+1) + 2(\alpha^2-5\alpha) + 1\left(\dfrac{\alpha-3}{2}\right) = 0$

Multiplying by 2 to clear the denominator:
$6(\alpha+1) + 4(\alpha^2-5\alpha) + (\alpha-3) = 0$

$6\alpha + 6 + 4\alpha^2 - 20\alpha + \alpha - 3 = 0$

$4\alpha^2 - 13\alpha + 3 = 0$

Factoring the quadratic equation:
$4\alpha^2 - 12\alpha - \alpha + 3 = 0$

$4\alpha(\alpha - 3) - 1(\alpha - 3) = 0$

$(4\alpha - 1)(\alpha - 3) = 0 \Rightarrow \alpha = \dfrac{1}{4} \text{ or } \alpha = 3$

For $P'$ to be the image of $P$, both the midpoint and perpendicularity conditions must hold simultaneously. The only common value satisfying both conditions is $\alpha = 3$.

Answer: Only $3$