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JEE Main 2021 · Mathematics · Three Dimensional Geometry · hard

The plane through the line of intersection of x+y+z=1 and 2x+3y-z+4=0 that also passes through the point (1,1,1) is at what perpendicular distance…

Single correct option

The plane through the line of intersection of x+y+z=1x+y+z=1 and 2x+3y−z+4=02x+3y-z+4=0 that also passes through the point (1,1,1)(1,1,1) is at what perpendicular distance from the origin?
  1. A. 826\dfrac{8}{\sqrt{26}}
  2. B. 430\dfrac{4}{\sqrt{30}}
  3. C. 85\dfrac85
  4. D. 830\dfrac{8}{\sqrt{30}}
Show answer and solution

Answer: (D) 830\dfrac{8}{\sqrt{30}}

Every plane through the line of intersection has the form (x+y+z−1)+λ(2x+3y−z+4)=0(x+y+z-1)+\lambda(2x+3y-z+4)=0. Substituting (1,1,1)(1,1,1) gives 2+8λ=02+8\lambda=0, so λ=−14\lambda=-\dfrac14. The plane is 4(x+y+z−1)−(2x+3y−z+4)=04(x+y+z-1)-(2x+3y-z+4)=0, i.e. 2x+y+5z−8=0.2x+y+5z-8=0. Its distance from the origin is ∣−8∣22+12+52=830.\frac{|{-8}|}{\sqrt{2^2+1^2+5^2}}=\frac{8}{\sqrt{30}}. A check: the point (1,1,1)(1,1,1) gives 2+1+5−8=02+1+5-8=0, as required. Any two points of the family give the same plane up to a scale factor, so the distance does not depend on the multiple used.
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Original question written in the JEE Main 2021 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake