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JEE Main 2022 · 27 Jul · Shift 1, Mathematics question 23

Let the line x−37=y−2−1=z−3−4\dfrac{x-3}7=\dfrac{y-2}{-1}=\dfrac{z-3}{-4} intersect the plane containing the lines x−41=y+1−2=z1\dfrac{x-4}1=\dfrac{y+1}{-2}=\dfrac z1 and 4ax−y+5z−7a=0=2x−5y−z−34ax-y+5z-7a=0=2x-5y-z-3, a∈Ra\in\mathbf R at the point P(α,β,γ)P(\alpha,\beta,\gamma). Then the value of α+β+γ\alpha+\beta+\gamma equals ______.

Answer and worked solution

Official answer: 12

Plane through line 2: (4ax−y+5z−7a)+μ(2x−5y−z−3)=0(4ax-y+5z-7a)+\mu(2x-5y-z-3)=0.

Contains line 1.

  • Through (4,−1,0)(4,-1,0): 9a+1+10μ=09a+1+10\mu=0.
  • Parallel to (1,−2,1)(1,-2,1): 4a+11μ+7=04a+11\mu+7=0.

So a=1a=1, μ=−1\mu=-1, and the plane is x+2y+3z=2x+2y+3z=2.

Intersect with (3+7t,2−t,3−4t)(3+7t,2-t,3-4t): 16−7t=216-7t=2, t=2t=2, so P=(17,0,−5)P=(17,0,-5).

Answer: 12

Topics:Equation of a planeEquation of a line in space
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