Vectors & 3D GeometrymediumPYQ · JEE Main · 5 Apr 2026 · Shift 1 (Morning)Free

Square Distance Point to Line in 3D = 6 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
The square of the distance of the point P(5,6,7)P(5,6,7) from the line x−22=y−53=z−24\dfrac{x-2}{2}=\dfrac{y-5}{3}=\dfrac{z-2}{4} is
A33
B55
C66correct
D88
Solution
Step 1: Q=(2λ+2, 3λ+5, 4λ+2)Q=(2\lambda+2,\ 3\lambda+5,\ 4\lambda+2), d⃗=(2,3,4)\vec d=(2,3,4).
PQ→=Q−P=(2λ−3, 3λ−1, 4λ−5).\overrightarrow{PQ}=Q-P=(2\lambda-3,\ 3\lambda-1,\ 4\lambda-5).
 Step 2: PQ→⋅d⃗=0\overrightarrow{PQ}\cdot\vec d=0:
(2λ−3)(2)+(3λ−1)(3)+(4λ−5)(4)=0.(2\lambda-3)(2)+(3\lambda-1)(3)+(4\lambda-5)(4)=0.
⇒(4λ−6)+(9λ−3)+(16λ−20)=29λ−29=0⇒λ=1.\Rightarrow(4\lambda-6)+(9\lambda-3)+(16\lambda-20)=29\lambda-29=0\Rightarrow\lambda=1.
 Step 3: Q=(4,8,6)Q=(4,8,6):
PQ2=(5−4)2+(6−8)2+(7−6)2=1+4+1=6.PQ^2=(5-4)^2+(6-8)^2+(7-6)^2=1+4+1=6.
 Correct answer: (3)
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