Vectors & 3D GeometrymediumPYQ · JEE Main · 5 Apr 2026 · Shift 2 (Afternoon)Free

Vectors & 3D Geometry: Let Pqr Lie Distance Centroid Pqr Equals (JEE Main 2026)

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

Question

Let

\triangle PQR

be such that

P

and

Q

lie on

\dfrac{x+3}{8}=\dfrac{y-4}{2}=\dfrac{z+1}{2}

and are at distance 6 from

R(1,2,3)

. If

(\alpha,\beta,\gamma)

is the centroid of

\triangle PQR

, then

\alpha+\beta+\gamma

equals:

4

5

6

correct

8

Solution

Step 1: Parametrize points on the line A general point on the line is

(8\lambda-3,\ 2\lambda+4,\ 2\lambda-1)

. Step 2: Apply the distance condition from

R(1,2,3)

(8\lambda-4)^2+(2\lambda+2)^2+(2\lambda-4)^2 = 36

64\lambda^2-64\lambda+16+4\lambda^2+8\lambda+4+4\lambda^2-16\lambda+16 = 36

72\lambda^2-72\lambda = 0 \implies \lambda(\lambda-1)=0 \implies \lambda=0 \text{ or } \lambda=1

Step 3: Determine

P

and

Q

\lambda=0:\; P=(-3,4,-1); \qquad \lambda=1:\; Q=(5,6,1)

Step 4: Compute the centroid

G = \left(\frac{-3+5+1}{3},\ \frac{4+6+2}{3},\ \frac{-1+1+3}{3}\right) = (1,4,1)

\alpha+\beta+\gamma = 1+4+1 = 6

Answer: (3)

Still stuck on this question?Ask your doubt on WhatsApp

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.