If (2α+1, α^2-3α, α-1 2 ) is the image of (α, 2α, 1) in the line x-2 3= y-1 2= z 1, then the possible value(s) of α is (are)

Let the given point be $P = (\alpha, 2\alpha, 1)$P=(α,2α,1) and its image be $P' = \left(2\alpha+1, \alpha^2-3\alpha, \dfrac{\alpha-1}{2}\right)$P′=(2α+1,α2−3α,2α−1​).

For $P'$P′ to be the image of $P$P in the given line $L: \dfrac{x-2}{3}=\dfrac{y-1}{2}=\dfrac{z}{1}$L:3x−2​=2y−1​=1z​, two conditions must be satisfied:

1. The midpoint of $P$P and $P'$P′ must lie on the line $L$L.

2. The line segment $PP'$PP′ must be perpendicular to the line $L$L.

The midpoint $M$M of $P$P and $P'$P′ is:

$$M = \left( \dfrac{\alpha + 2\alpha + 1}{2}, \dfrac{2\alpha + \alpha^2 - 3\alpha}{2}, \dfrac{1 + \dfrac{\alpha-1}{2}}{2} \right) = \left( \dfrac{3\alpha + 1}{2}, \dfrac{\alpha^2 - \alpha}{2}, \dfrac{\alpha + 1}{4} \right)$$M=​2α+2α+1​,22α+α2−3α​,21+2α−1​​​=(23α+1​,2α2−α​,4α+1​)

Since $M$M lies on $L$L, its coordinates must satisfy the equation of the line:

$$\dfrac{\dfrac{3\alpha + 1}{2} - 2}{3} = \dfrac{\dfrac{\alpha^2 - \alpha}{2} - 1}{2} = \dfrac{\dfrac{\alpha + 1}{4}}{1}$$323α+1​−2​=22α2−α​−1​=14α+1​​

Simplifying the terms, we get:

$$\dfrac{\alpha - 1}{2} = \dfrac{\alpha^2 - \alpha - 2}{4} = \dfrac{\alpha + 1}{4}$$2α−1​=4α2−α−2​=4α+1​

Equating the first and third parts:

$$\dfrac{\alpha - 1}{2} = \dfrac{\alpha + 1}{4}$$2α−1​=4α+1​

$$2(\alpha - 1) = \alpha + 1 \Rightarrow 2\alpha - 2 = \alpha + 1 \Rightarrow \alpha = 3$$2(α−1)=α+1⇒2α−2=α+1⇒α=3

Substituting $\alpha = 3$α=3 into the middle part to verify:

$$\dfrac{3^2 - 3 - 2}{4} = \dfrac{4}{4} = 1$$432−3−2​=44​=1

The first and third parts also equal $1$1, so $\alpha = 3$α=3 satisfies the midpoint condition.

Now, let us check the perpendicularity condition. The direction ratios of $\overrightarrow{PP'}$PP′ are:

$$\left(2\alpha+1 - \alpha, \alpha^2-3\alpha - 2\alpha, \dfrac{\alpha-1}{2} - 1\right) = \left(\alpha+1, \alpha^2-5\alpha, \dfrac{\alpha-3}{2}\right)$$(2α+1−α,α2−3α−2α,2α−1​−1)=(α+1,α2−5α,2α−3​)

The direction ratios of the line $L$L are $(3, 2, 1)$(3,2,1). The dot product must be zero:

$$3(\alpha+1) + 2(\alpha^2-5\alpha) + 1\left(\dfrac{\alpha-3}{2}\right) = 0$$3(α+1)+2(α2−5α)+1(2α−3​)=0

Multiplying by 2 to clear the denominator:

$$6(\alpha+1) + 4(\alpha^2-5\alpha) + (\alpha-3) = 0$$6(α+1)+4(α2−5α)+(α−3)=0

$$6\alpha + 6 + 4\alpha^2 - 20\alpha + \alpha - 3 = 0$$6α+6+4α2−20α+α−3=0

$$4\alpha^2 - 13\alpha + 3 = 0$$4α2−13α+3=0

Factoring the quadratic equation:

$$4\alpha^2 - 12\alpha - \alpha + 3 = 0$$4α2−12α−α+3=0

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

$$(4\alpha - 1)(\alpha - 3) = 0 \Rightarrow \alpha = \dfrac{1}{4} \text{ or } \alpha = 3$$(4α−1)(α−3)=0⇒α=41​ or α=3

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

Answer: Only $3$3