Let the point A be the foot of perpendicular drawn from the point \(\mathrm{P}(\mathrm{a}, \mathrm{b}, 0)\) on the line \(\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-\alpha}{3}\). If the midpoint of the line segment PA is \(\left(0, \frac{3}{4}, \frac{-1}{4}\right)\), then the value of \(\mathrm{a}^{2}+\mathrm{b}^{2} +\alpha^{2}\) is equal to :

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Three Dimensional Geometry

Let the point A be the foot of perpendicular drawn from the point

P (a, b, 0)

on the line

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

. If the midpoint of the line segment PA is

(0, \frac{3}{4}, \frac{- 1}{4})

, then the value of

a^{2} + b^{2} + α^{2}

is equal to :

Answer: (A)

A general point of the line is

A = (1 + 2 t, 2 + t, α + 3 t)

. Equating the midpoint of

P A

(0, \frac{3}{4}, - \frac{1}{4})

gives

a = - 1 - 2 t, b = - \frac{1}{2} - t, α = - \frac{1}{2} - 3 t

. The vector

P A = (2 + 4 t, \frac{5}{2} + 2 t, α + 3 t)

must be perpendicular to the direction

(2, 1, 3)

2 (2 + 4 t) + (\frac{5}{2} + 2 t) + 3 (α + 3 t) = 0.

Substituting

α = - \frac{1}{2} - 3 t

gives

10 t + 5 = 0

, so

t = - \frac{1}{2}

, and then

a = 0, b = 0, α = 1

. Therefore

a^{2} + b^{2} + α^{2} = 1

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Single correct option