Q6 / 105 · NTA Abhyas (JEE Main) 2020 · Mathematics · Three Dimensional Geometry

The perpendicular bisector of a line segment with end points 1,2,6 and -3,6,2 passes through -6,2,4 and has the equation of the form x+6l=y-2m=z-4n (Where l,m,n are integers, l is a prime number and l>0), then the value of lmn-l+m+n equals to

  • A
    -3
  • B
    -5
  • Correct answer
    C
    -7
  • D
    -9

Solution

Midpoint of the line segment is 1-31,2+62,6+22≡-1,4,4
Parallel vector to the required line =-1+6i^+4-2j^+4-4k^
=5i^+2j^+0k^
Hence, equation of the line is
x+65-y-22=z-40
⇒l=5, m=2, n=0

Hence, the correct answer is option: 'C'.

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