Let the point A be the foot of perpendicular drawn from the point P(a, b, 0) on the line x - 1/2 = y - 2/1 = z - α/3 . If the midpoint of the line segment PA is (0, 3/4, -1/4 ) , then the value of a 2 + b 2 + α 2 is equal to :

Mathematics · 3D Geometry

JEE Main 2026 — 2 April, Evening Shift — Question 37

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

Option A:
1
Correct
Option B:
2
Option C:
6
Option D:
9

Answer: A

Step-by-step solution

$\frac{2 \mathrm{r}+1+\mathrm{a}}{2}=0 \Rightarrow 2 \mathrm{r}+\mathrm{a}=-1$ $\frac{\mathrm{r}+2+\mathrm{b}}{2}=\frac{3}{4} \Rightarrow \mathrm{r}+\mathrm{b}=-\frac{1}{2}$ $\frac{3 r+\alpha}{2}=\frac{-1}{4} \Rightarrow 3 r+\alpha=\frac{-1}{2}$ $2 \cdot \mathrm{a}+1 \cdot(\mathrm{~b}-3 / 4)+3 \cdot \frac{1}{4}=0 \Rightarrow \mathrm{a}+\mathrm{b}=0$ $\Rightarrow 2(-1-2 \mathrm{r})+(-\mathrm{r}-1 / 2)=0$ $\Rightarrow-3 / 2=3 \mathrm{r}=1 \Rightarrow \mathrm{r}=-1 / 2, \mathrm{a}=0, \mathrm{~b}=0, \alpha=1$ $\therefore \mathrm{a}^{2}+\mathrm{b}^{2}+\alpha^{2}=1$

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Exam: JEE Main 2026
Paper: 2 April, Evening Shift
Subject: Mathematics
Chapter: 3D Geometry
Topic: Straight Lines in 3D Geometry