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MathsJEE Main 6 Apr 2026, Shift 1·Q15·MCQ·Tough

3D Geometry · Image of point in line

Let the image of the point P(1,6,a)P(1, 6, a) in the line L:x1=y−12=z−a+1bL : \dfrac{x}{1} = \dfrac{y - 1}{2} = \dfrac{z - a + 1}{b}, b>0b > 0, be (a3,0,a+c)\left(\dfrac{a}{3}, 0, a + c\right). If S(α,β,γ)S(\alpha, \beta, \gamma), α>0\alpha > 0, is the point on LL such that the distance of SS from the foot of perpendicular from the point PP on LL is 2142\sqrt{14}, then α+β+γ\alpha + \beta + \gamma is equal to:
  1. (1)19
  2. (2)20
  3. (3)21
  4. (4)22

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