If the shortest distance between the straight lines $3(x-1)=6(y-2)=2(z-1)$ and $4(x-2)=2(y-\lambda)=(z-3)$, $\lambda\in\mathbf{R}$ is $\frac{1}{\sqrt{38}}$, then the integral value of $\lambda$ is equal to :

JEE Main 2021 · MathsMultiple choiceSingle correctMediumCalculation

JEE Main 22 July 2021, Shift 2, Maths Q63

Question 63 of 90 in this shift, Maths question 3 of 30, Section A.

If the shortest distance between the straight lines

3(x-1)=6(y-2)=2(z-1)

and

4(x-2)=2(y-\lambda)=(z-3)

\lambda\in\mathbf{R}

\frac{1}{\sqrt{38}}

, then the integral value of

\lambda

is equal to :

(1) $-1$
(2) $2$
(3) $3$ Official answer
(4) $5$

Official answer

Option 3

NTA final key.

Chapter: Three Dimensional Geometry
Topic: Skew Lines
Idea tested: Finding an Unknown Parameter from Shortest Distance

Same idea in other shifts

Asked 3× in all

8 Apr 2025, Shift 2 · Q15Let the values of $\lambda$ for which the shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and…HardSingle correct
6 Apr 2024, Shift 2 · Q28If the shortest distance between the lines $\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}$ and…MediumNumerical value

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.