If the shortest distance between the lines $\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}$ is $\frac{44}{\sqrt{30}}$, then the largest possible value of $|\lambda|$ is equal to ________.

JEE Main 2024 · MathsNumerical answerNumerical valueMediumCalculation

JEE Main 6 April 2024, Shift 2, Maths Q28

Question 28 of 90 in this shift, Maths question 28 of 30, Section B.

If the shortest distance between the lines

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

and

\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}

\frac{44}{\sqrt{30}}

, then the largest possible value of

|\lambda|

is equal to ________.

Official answer

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
22 Jul 2021, Shift 2 · Q63If 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…MediumSingle correct

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.