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

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

Let L be the line (x+1)/(2)=(y+1)/(3)=(z+3)/(6) and let S be the...

Let L be the line x+12=y+13=z+36\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6} and let S be the set of all points (a,b,c)(\mathrm{a}, \mathrm{b}, \mathrm{c}) on L, whose distance from the line x+12=y+13=z−90\frac{x+1}{2}=\frac{y+1}{3}=\frac{z-9}{0} along the line LL is 7. Then ∑(a,b,c)∈S(a+b+c)\sum_{(a, b, c) \in S}(a+b+c) is equal to :

Options

  1. A6
  2. B34
  3. C40
  4. D28

Correct answer

B. 34

Step-by-step solution

M is the point of intersection of L1L_1 & L2L_2
⇒2λ−1=2μ−1,3λ−1=3μ−1,6λ−3=9\Rightarrow 2\lambda - 1 = 2\mu - 1, 3\lambda - 1 = 3\mu - 1, 6\lambda - 3 = 9
⇒λ=2=μ\Rightarrow \lambda = 2 = \mu
⇒M(3,5,9)\Rightarrow M(3, 5, 9)
Now let point P be (2K−1,3K−1,6K−3)(2K - 1, 3K - 1, 6K - 3) on L2L_2
such that PM=7PM = 7
⇒(2K−4)2+(3K−6)2+(6K−12)2=7\Rightarrow \sqrt{(2K - 4)^2 + (3K - 6)^2 + (6K - 12)^2} = 7
⇒49K2+196−196K=49\Rightarrow 49K^2 + 196 - 196K = 49
⇒K2+4−4K=1\Rightarrow K^2 + 4 - 4K = 1
⇒K2−4K+3=0\Rightarrow K^2 - 4K + 3 = 0
⇒K=1,3\Rightarrow K = 1, 3
So points P & Q are (1,2,3)(1, 2, 3) & (5,8,15)(5, 8, 15)
So sum of all co-ordinates of P & Q =34= 34
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