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Mathematics · 3D Geometry

JEE Main 2026 — 5 April, Evening Shift — Question 39

If the distance of the point (a,2,5)(a,2,5) from the image of the point (1,2,7)(1,2,7) in the line x−11=y−21=z−72\frac{x-1}{1} = \frac{y-2}{1} = \frac{z-7}{2} is 4,4, then the sum of all possible values of aa is equal to:

  1. Option A:

    1111

  2. Option B:

    99

  3. Option C:

    66

    Correct
  4. Option D:

    44

Answer: C

Step-by-step solution

PQ⊥L⇒(α−1)+(β−2)+2(γ−7)=0P Q \perp L \Rightarrow(\alpha-1)+(\beta-2)+2(\gamma-7)=0

\Rightarrow \alpha+\beta+2 \gamma=17 \end{gathered}$$ M is mid point of PQ which will satisfy L $\frac{\alpha+1}{2}=\frac{\frac{\beta+2}{2}-1}{1}=\frac{\frac{\gamma+7}{2}-2}{2}$ $\Rightarrow \frac{\alpha+1}{2}=\frac{\beta}{2}=\frac{\gamma+3}{4}$ $$\begin{gathered} \Rightarrow \alpha+1=\beta \end{gathered}$$ and $2 \beta=\gamma+3$ $$\begin{gathered} \Rightarrow \alpha=3, \beta=4, \gamma=5 \end{gathered}$$ Distance from (a, 2, 5) is $=\sqrt{(a-3)^{2}+4+0}=4$ $(\mathrm{a}-3)^{2}+4=16 \Rightarrow \mathrm{a}^{2}-6 \mathrm{a}-3=0 \Rightarrow$ sum of values of $\mathrm{a}=6$
Solution figure

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Exam
JEE Main 2026
Paper
5 April, Evening Shift
Subject
Mathematics
Chapter
3D Geometry
Topic
Straight Lines in 3D Geometry

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