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JEE Main 2024 · Mathematics · Three Dimensional Geometry · hard

A plane contains the line (x-1)/(2)=(y+1)/(3)=(z-2)/(6) and the point (7,1,-1). If (α,β,γ) is the foot of the perpendicular drawn from the origin to…

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A plane contains the line x−12=y+13=z−26\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{6} and the point (7,1,−1)(7,1,-1). If (α,β,γ)(\alpha,\beta,\gamma) is the foot of the perpendicular drawn from the origin to this plane, then α+β+γ\alpha+\beta+\gamma equals
  1. A. −137-\dfrac{13}{7}
  2. B. 1349\dfrac{13}{49}
  3. C. −1349-\dfrac{13}{49}
  4. D. −2649-\dfrac{26}{49}
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Answer: (C) −1349-\dfrac{13}{49}

The line passes through A(1,−1,2)A(1,-1,2) with direction d⃗=(2,3,6)\vec d=(2,3,6). For P(7,1,−1)P(7,1,-1), AP→=(6,2,−3)\overrightarrow{AP}=(6,2,-3). A normal is d⃗×AP→=(−21,42,−14)=−7 (3,−6,2)\vec d\times\overrightarrow{AP}=(-21,42,-14)=-7\,(3,-6,2). So the plane is 3x−6y+2z=k3x-6y+2z=k; using AA, k=3+6+4=13k=3+6+4=13 (and PP gives 21−6−2=1321-6-2=13). The foot of the perpendicular from the origin is λ(3,−6,2)\lambda(3,-6,2) with λ⋅49=13\lambda\cdot49=13, so (α,β,γ)=1349(3,−6,2)(\alpha,\beta,\gamma)=\dfrac{13}{49}(3,-6,2) and α+β+γ=1349(−1)=−1349\alpha+\beta+\gamma=\dfrac{13}{49}(-1)=-\dfrac{13}{49}. The value 137\dfrac{13}{7} is the distance, not the coordinate sum. The point (7,1,−1)(7,1,-1) was chosen so that AP→⋅d⃗=12+6−18=0\overrightarrow{AP}\cdot\vec d=12+6-18=0, which keeps the normal's length equal to ∣d⃗∣∣AP→∣=49|\vec d||\overrightarrow{AP}|=49 and the arithmetic clean.
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