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← JEE Main 2022 · 25 Jul · Shift 1
MathematicsInteger+4 / -1Question 30 of 90

JEE Main 2022 · 25 Jul · Shift 1, Mathematics question 30

The line of shortest distance between the lines x−20=y−11=z1\dfrac{x-2}0=\dfrac{y-1}1=\dfrac z1 and x−32=y−52=z−11\dfrac{x-3}2=\dfrac{y-5}2=\dfrac{z-1}1 makes an angle of cos⁡−1(227)\cos^{-1}\left(\sqrt{\dfrac2{27}}\right) with the plane P:ax−y−z=0P:ax-y-z=0, (a>0)(a>0). If the image of the point (1,1,−5)(1,1,-5) in the plane P is (α,β,γ)(\alpha,\beta,\gamma), then α+β−γ\alpha+\beta-\gamma is equal to ______.

Answer and worked solution

Official answer: Dropped: full marks to everyone

This question was dropped in NTA's final key.

Why. The shortest-distance line has direction (0,1,1)×(2,2,1)=(−1,2,−2)(0,1,1)\times(2,2,1)=(-1,2,-2). With normal (a,−1,−1)(a,-1,-1) the sine of the angle is a3a2+2\dfrac{a}{3\sqrt{a^2+2}}, but cos⁡−1227\cos^{-1}\sqrt{\frac2{27}} needs sine2=2527^2=\frac{25}{27}, i.e. 27a2=225a2+45027a^2=225a^2+450: no real aa.