Verified Solution • 2025–26 Session moderate

QJEEM25Maeb84: JEE Main 2025 Mathematics - Class 12 JEEM25Maeb84 Solution: Let a straight line pass through the point and be perpendicular …

Let a straight line LL pass through the point P(2,−1,3)P(2, -1, 3) and be perpendicular to the lines x−12=y+11=z−3−2\frac{x - 1}{2} = \frac{y + 1}{1} = \frac{z - 3}{-2} and x−31=y−23=z+24\frac{x - 3}{1} = \frac{y - 2}{3} = \frac{z + 2}{4} . If the line LL intersects the yzyz -plane at the point QQ , then the distance between the points PP and QQ is:

Multiple Choice Options

  • A 10\sqrt{10}
  • B 22
  • C 232\sqrt{3}
  • D 33

Step-by-Step Verified Solution

Vector parallel to 'L'

=|i^j^k^21−2134|=10i^−10j^+5k^=5(2i^−2j^+k^)

Equation of 'L'

x−22=y+1−2=z−31=λ( say )

Let Q(2λ+2,−2λ−1,λ+3)\mathrm{Q}(2 \lambda+2,-2 \lambda-1, \lambda+3)

⇒2λ+2=0⇒λ=−1⇒Q(0,1,2)d(P,Q)=3

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