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

(C) Only $3, \frac{1}{4}$ and -1

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Three Dimensional Geometry

(2 α + 1, α^{2} - 3 α, \frac{α - 1}{2})

is the image of

(α, 2 α, 1)

in the line

\frac{x - 2}{3} = \frac{y - 1}{2} = \frac{z}{1}

, then the possible value(s) of

α

is(are) :

Answer: (A)

Let

A = (α, 2 α, 1)

and

B = (2 α + 1, α^{2} - 3 α, \frac{α - 1}{2})

. If

B

is the mirror image of

A

in the line

\frac{x - 2}{3} = \frac{y - 1}{2} = \frac{z}{1}

, then

A B

is perpendicular to the direction

(3, 2, 1)

and the midpoint of

A B

lies on the line. The vector

A B = (α + 1, α^{2} - 5 α, \frac{α - 3}{2})

; the perpendicularity

3 (α + 1) + 2 (α^{2} - 5 α) + \frac{α - 3}{2} = 0

reduces to

4 α^{2} - 13 α + 3 = 0

, i.e.

α = 3

α = \frac{1}{4}

. Imposing that the midpoint

(\frac{3 α + 1}{2}, \frac{α ^{2} - α}{2}, \frac{α + 1}{4})

lies on the line gives, from the

x

and

z

relation,

2 (3 α + 1) = 3 α + 11 \Rightarrow α = 3

, and from the

y

and

z

relation

α^{2} - 2 α - 3 = 0 \Rightarrow α = 3

- 1

. The only value satisfying all conditions is

α = 3

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Single correct option