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JEE Main 202622 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2026 · 22 January · Morning Shift · Mathematics · Three Dimensional Geometry

If the image of the point P(1,2, a) in the line...

If the image of the point P(1,2,a)\mathrm{P}(1,2, a) in the line x−63=y−72=7−z2\frac{x-6}{3}=\frac{y-7}{2}=\frac{7-\mathrm{z}}{2} is Q(5,b,c)\mathrm{Q}(5, b, \mathrm{c}), then a2+b2+c2a^{2}+b^{2}+c^{2} is equal to

Options

  1. A298
  2. B293
  3. C264
  4. D283

Correct answer

A. 298

Step-by-step solution

Line: point (6,7,7)(6,7,7), direction (3,2,−2)(3,2,-2).
General point on line: (6+3t,7+2t,7−2t)(6+3t, 7+2t, 7-2t).
Foot of perpendicular from P(1,2,a)P(1,2,a): PM⃗⋅(3,2,−2)=0\vec{PM} \cdot (3,2,-2) = 0.
3(5+3t)+2(5+2t)−2(7−2t−a)=03(5+3t)+2(5+2t)-2(7-2t-a) = 0
⇒17t+11+2a=0\Rightarrow 17t+11+2a = 0.
Image Q=2M−P=(11+6t,12+4t,14−4t−a)Q = 2M - P = (11+6t, 12+4t, 14-4t-a).
Given Qx=5Q_x = 5: 11+6t=5⇒t=−111+6t = 5 \Rightarrow t = -1.
From 17(−1)+11+2a=0⇒a=317(-1)+11+2a = 0 \Rightarrow a = 3.
b=12+4(−1)=8b = 12+4(-1) = 8, c=14−4(−1)−3=15c = 14-4(-1)-3 = 15.
a2+b2+c2=9+64+225=298a^2+b^2+c^2 = 9+64+225 = 298.
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