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Friday, 9 October

JEE Main 2026 · MathsMultiple choiceSingle correctMediumCalculation

JEE Main 4 April 2026, Shift 1, Maths Q15

Question 15 of 75 in this shift, Maths question 15 of 25, Section A.

The square of the distance of the point (−2,−8,6)(-2,-8,6) from the line x−11=y−12=z−1\frac{x-1}{1}=\frac{y-1}{2}=\frac{z}{-1} along the line x+51=y+5−1=z2\frac{x+5}{1}=\frac{y+5}{-1}=\frac{z}{2} is equal to:
  1. (1)33
  2. (2)66Official answer
  3. (3)88
  4. (4)1212

Official answer

Option 2

NTA final key.

Same idea in other shifts

Asked 2× in all
  1. 16 Mar 2021, Shift 2 · Q83If the distance of the point (1, −2-2, 3) from the plane x+2y−3z+10=0x+2y-3z+10=0 measured parallel to the line,…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 42 of the 174 JEE Main shifts held from 2017 to 2026.