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JEE Main 2020 · Mathematics · Three Dimensional Geometry · hard

Let Q be the image (mirror reflection) of the point P(1,2,3) in the line x/1=(y-1)/(2)=z/2. The distance of Q from the origin is

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Let QQ be the image (mirror reflection) of the point P(1,2,3)P(1,2,3) in the line x1=y−12=z2\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z}{2}. The distance of QQ from the origin is
  1. A. 14\sqrt{14}
  2. B. 17\sqrt{17}
  3. C. 323\sqrt2
  4. D. 44
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Answer: (C) 323\sqrt2

A general point of the line is F(λ, 1+2λ, 2λ)F(\lambda,\,1+2\lambda,\,2\lambda) with direction (1,2,2)(1,2,2). The foot of the perpendicular from PP satisfies PF⃗⋅(1,2,2)=0\vec{PF}\cdot(1,2,2)=0: (λ−1)+2(2λ−1)+2(2λ−3)=0⇒9λ=9⇒λ=1,(\lambda-1)+2(2\lambda-1)+2(2\lambda-3)=0\Rightarrow 9\lambda=9\Rightarrow\lambda=1, so F=(1,3,2)F=(1,3,2). The image is Q=2F−P=(1,4,1)Q=2F-P=(1,4,1) because FF is the midpoint of PQPQ. Then ∣OQ∣=1+16+1=32|OQ|=\sqrt{1+16+1}=3\sqrt2. The distractor 14\sqrt{14} is ∣OF∣|OF|, the foot instead of the image.
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