Let Q ( a , b , c ) be the image of the point P (3,2,1) in the line x-1/1=y/2=z-1/1 . Then the distance of Q from the line x-9/3=y-9/2=z-5/-2 is

Mathematics · Straight lines

JEE Main 2026 — 28 January, Evening Shift — Question 16

Let $\mathrm{Q}(\mathrm{a}, \mathrm{b}, \mathrm{c})$ be the image of the point $\mathrm{P}(3,2,1)$ in the line $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$ . Then the distance of $Q$ from the line $\frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2}$ is

Option A:
6
Option B:
8
Option C:
7
Correct
Option D:
5

Answer: C

Step-by-step solution

drs of $\mathrm{PN}=\langle\mathrm{r}-2,2 \mathrm{r}-2, \mathrm{r}\rangle$

$1 \cdot(\mathrm{r}-2)+2(2 \mathrm{r}-2)+1 \cdot(\mathrm{r})=0$

$6 \mathrm{r}=6 \Rightarrow \mathrm{r}=1$

$\therefore \mathrm{N} \equiv(2,2,2)$

$\Rightarrow \mathrm{Q} \equiv(1,2,3)$

$\mathrm{AQ}=\sqrt{64+49+4}=\sqrt{117}$

$\mathrm{AM}=\left|\frac{24+14-4}{\sqrt{9+4+4}}\right|=\frac{34}{\sqrt{17}}=2 \sqrt{17}$

$\therefore \mathrm{QM}=\sqrt{117-68}=\sqrt{49}=7$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 28 January, Evening Shift
Subject: Mathematics
Chapter: Straight lines
Topic: Angle between lines, perpendicular distance & distance between parallel lines, foot, image