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JEE Main 202628 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
28 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2026 · 28 January · Morning Shift · Mathematics · Three Dimensional Geometry

If the distances of the point (1,2, a) from the line...

If the distances of the point (1,2,a)(1,2, a) from the line x−11=y2=z−11\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1} along the lines L1:x−13=y−24=z−ab\mathrm{L}_{1}: \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b} and L2:x−11=y−24=z−ac\mathrm{L}_{2}: \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c} are equal, then a+b+ca+b+c is equal to

Options

  1. A5
  2. B7
  3. C4
  4. D6

Correct answer

B. 7

Step-by-step solution

Line LL: x−11=y2=z−11\frac{x-1}{1} = \frac{y}{2} = \frac{z-1}{1}

L1L_1: x−13=y−24=z−ab=λ\frac{x-1}{3} = \frac{y-2}{4} = \frac{z-a}{b} = \lambda

L2L_2: x−11=y−24=z−ac=μ\frac{x-1}{1} = \frac{y-2}{4} = \frac{z-a}{c} = \mu

Point A(3λ+1,4λ+2,bλ+a)A(3\lambda+1, 4\lambda+2, b\lambda+a) lies on LL:
3λ1=4λ+22=bλ+a−11\frac{3\lambda}{1} = \frac{4\lambda+2}{2} = \frac{b\lambda+a-1}{1}

λ=1\lambda = 1, a+b−1=3⇒a+b=4a + b - 1 = 3 \Rightarrow a + b = 4 ...(1), A=(4,6,4)A = (4, 6, 4)

Point B(μ+1,4μ+2,cμ+a)B(\mu+1, 4\mu+2, c\mu+a) lies on LL:
2μ=4μ+2⇒μ=−12\mu = 4\mu+2 \Rightarrow \mu = -1, a−c−1=−1⇒a=ca - c - 1 = -1 \Rightarrow a = c ...(2), B=(0,−2,0)B = (0, -2, 0)

Since PA=PBPA = PB where P=(1,2,a)P = (1, 2, a):
9+16+(a−4)2=1+16+a29 + 16 + (a-4)^2 = 1 + 16 + a^2
⇒a=3\Rightarrow a = 3, so c=3c = 3, b=1b = 1

a+b+c=7a + b + c = 7
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