JEE Main 2021MathematicsThree Dimensional GeometryEasyNumerical

JEE Main 2021 — Three Dimensional Geometry Question with Solution

JEE Main 2021 (31 Aug Shift 1)

Question

The square of the distance of the point of intersection of the line x-12=y-23=z+16 and the plane 2 x-y+z=6 from the point (-1,-1,2) is

Enter your answer

Show full solutionCorrect answer: 61
Correct answer
61

Step-by-step explanation

Equation of line

x-12=y-23=z+16

Let x-12=y-23=z+16=λ

⇒x = 2λ+1, y = 3λ+2, z=6λ-1

⇒   Let P(2λ+1,3λ+2,6λ-1) is point of intersection of line and plane

So it should satisfy the given plane.

⇒  2(2λ+1)-3λ-2+6λ-1=6

7λ=7⇒λ=1

Point P(3,5,5)

Square of distance from point (-1,-1,2)  and P(3,5,5)

=x2-x12 +y2-y12 +z2-z12  

=42+62+32=61

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Three Dimensional Geometry chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2021, covering the Three Dimensional Geometry chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.