JEE MainMathematics

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Centroid of a Triangle Formed by Line Points Equidistant from a Given Point

Let P and Q be the points on the line (x+3)/8 = (y-4)/2 = (z+1)/2 which are at a distance of 6 units from the point R(1,2,3). If the centroid of the triangle PQR is (α, β, γ), then α^2 + β^2 + γ^2 is:

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A26B36C18D24

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Correct answer

The centroid of triangle $P Q R$ is $(α, β, γ) = (1, 4, 1)$ , giving $α^{2} + β^{2} + γ^{2} = 1^{2} + 4^{2} + 1^{2} = 18$ .

Option analysis

Why each option works or fails

A · 26

Calculated the sum of squares of the midpoint of $P Q$ instead of the centroid of triangle $P Q R$ . Remember that the centroid coordinates of $△ P Q R$ involve averaging the coordinates of all three points $P$ , $Q$ , and $R$ , not just the midpoint of $P Q$ .

B · 36

Made an algebraic sign error when expanding the distance equation, leading to incorrect parameter values for $P$ and $Q$ . Carefully expand $(8 λ - 4)^{2} + (2 λ + 2)^{2} + (2 λ - 4)^{2} = 36$ to find the roots $λ = 0, 1$ with the correct signs.

C · 18

The solution correctly identifies the coordinates of $P$ and $Q$ , evaluates the centroid of $△ P Q R$ , and sums the squares of its components.

D · 24

Arithmetically miscalculated the sum of squares $1^{2} + 4^{2} + 1^{2}$ or made an arithmetic error when averaging the coordinates. Double-check the arithmetic: $α = 1, β = 4, γ = 1 ⟹ 1^{2} + 4^{2} + 1^{2} = 1 + 16 + 1 = 18$ .

Reviewed route

Solution

StepWorking

01Given

Line equation: $\frac{x+3}{8} = rac{y-4}{2} = rac{z+1}{2} = \lambda$ . Point $R (1, 2, 3)$ . Points $P, Q$ lie on the line such that $P R = Q R = 6$ .

02Goal

Find the centroid $(α, β, γ)$ of $Δ P Q R$ and compute $α^{2} + β^{2} + γ^{2}$ .

03Approach

Express any point on the line in terms of parameter $λ$ as $(8 λ - 3, 2 λ + 4, 2 λ - 1)$ . Apply the distance formula to point $R (1, 2, 3)$ and set the square of the distance to $36$ . Solve the resulting quadratic for the two values of $λ$ , find coordinates of $P$ and $Q$ , compute the centroid of $Δ P Q R$ , and evaluate $α^{2} + β^{2} + γ^{2}$ .

04Execute

Using the distance formula from $(8 λ - 3, 2 λ + 4, 2 λ - 1)$ to $(1, 2, 3)$ : $(8 λ - 4)^{2} + (2 λ + 2)^{2} + (2 λ - 4)^{2} = 36$ Expanding the terms: $(64 λ^{2} - 64 λ + 16) + (4 λ^{2} + 8 λ + 4) + (4 λ^{2} - 16 λ + 16) = 36$ $72 λ^{2} - 72 λ + 36 = 36 ⟹ 72 λ (λ - 1) = 0$ Thus, $λ = 0$ or $λ = 1$ . For $λ = 0$ , $P = (- 3, 4, - 1)$ . For $λ = 1$ , $Q = (5, 6, 1)$ .

05Execute

Compute centroid $(α, β, γ)$ of $Δ P Q R$ with vertices $P (- 3, 4, - 1)$ , $Q (5, 6, 1)$ , $R (1, 2, 3)$ : $α = \frac{- 3 + 5 + 1}{3} = 1$ $β = \frac{4 + 6 + 2}{3} = 4$ $γ = \frac{- 1 + 1 + 3}{3} = 1$ Now calculate $α^{2} + β^{2} + γ^{2} = 1^{2} + 4^{2} + 1^{2} = 1 + 16 + 1 = 18$ .

✓Verify

Verify distance: $P R^{2} = (- 3 - 1)^{2} + (4 - 2)^{2} + (- 1 - 3)^{2} = 16 + 4 + 16 = 36 ⟹ P R = 6$ . $Q R^{2} = (5 - 1)^{2} + (6 - 2)^{2} + (1 - 3)^{2} = 16 + 16 + 4 = 36 ⟹ Q R = 6$ . Both points are verified correctly.

Hints that build this answer step by step

What is the parametric form of a general point $T (λ)$ on the line $\frac{x + 3}{8} = \frac{y - 4}{2} = \frac{z + 1}{2}$ ?
$(8 λ - 3, 2 λ + 4, 2 λ - 1)$
Using the distance condition $∣ T (λ) - R ∣ = 6$ where $R = (1, 2, 3)$ , what are the values of $λ$ corresponding to $P$ and $Q$ ?
$λ = 0$ and $λ = 1$
Given $P = (- 3, 4, - 1)$ (for $λ = 0$ ), $Q = (5, 6, 1)$ (for $λ = 1$ ), and $R = (1, 2, 3)$ , what are the centroid coordinates $(α, β, γ)$ and the value of $α^{2} + β^{2} + γ^{2}$ ?
$(α, β, γ) = (1, 4, 1)$ and $α^{2} + β^{2} + γ^{2} = 18$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 8 September 2026

Quick checks

Students also ask

Why do the two roots of the quadratic give points P and Q directly?

Both P and Q lie on the same line. They are also at the exact same distance of 6 units from R. Therefore, the single quadratic in parameter $λ$ must yield precisely the two points.

Answer

The centroid of triangle $P Q R$ is $(α, β, γ) = (1, 4, 1)$ , giving $α^{2} + β^{2} + γ^{2} = 1^{2} + 4^{2} + 1^{2} = 18$ .

Why each option works or fails

A: 26 - Calculated the sum of squares of the midpoint of $P Q$ instead of the centroid of triangle $P Q R$ . Remember that the centroid coordinates of $△ P Q R$ involve averaging the coordinates of all three points $P$ , $Q$ , and $R$ , not just the midpoint of $P Q$ .
B: 36 - Made an algebraic sign error when expanding the distance equation, leading to incorrect parameter values for $P$ and $Q$ . Carefully expand $(8 λ - 4)^{2} + (2 λ + 2)^{2} + (2 λ - 4)^{2} = 36$ to find the roots $λ = 0, 1$ with the correct signs.
C · correct: 18 - The solution correctly identifies the coordinates of $P$ and $Q$ , evaluates the centroid of $△ P Q R$ , and sums the squares of its components.
D: 24 - Arithmetically miscalculated the sum of squares $1^{2} + 4^{2} + 1^{2}$ or made an arithmetic error when averaging the coordinates. Double-check the arithmetic: $α = 1, β = 4, γ = 1 ⟹ 1^{2} + 4^{2} + 1^{2} = 1 + 16 + 1 = 18$ .

Step-by-step solution

Given: Line equation: $\frac{x+3}{8} = rac{y-4}{2} = rac{z+1}{2} = \lambda$ . Point $R (1, 2, 3)$ . Points $P, Q$ lie on the line such that $P R = Q R = 6$ .
Goal: Find the centroid $(α, β, γ)$ of $Δ P Q R$ and compute $α^{2} + β^{2} + γ^{2}$ .
Approach: Express any point on the line in terms of parameter $λ$ as $(8 λ - 3, 2 λ + 4, 2 λ - 1)$ . Apply the distance formula to point $R (1, 2, 3)$ and set the square of the distance to $36$ . Solve the resulting quadratic for the two values of $λ$ , find coordinates of $P$ and $Q$ , compute the centroid of $Δ P Q R$ , and evaluate $α^{2} + β^{2} + γ^{2}$ .
Execute: Using the distance formula from $(8 λ - 3, 2 λ + 4, 2 λ - 1)$ to $(1, 2, 3)$ : $(8 λ - 4)^{2} + (2 λ + 2)^{2} + (2 λ - 4)^{2} = 36$ Expanding the terms: $(64 λ^{2} - 64 λ + 16) + (4 λ^{2} + 8 λ + 4) + (4 λ^{2} - 16 λ + 16) = 36$ $72 λ^{2} - 72 λ + 36 = 36 ⟹ 72 λ (λ - 1) = 0$ Thus, $λ = 0$ or $λ = 1$ . For $λ = 0$ , $P = (- 3, 4, - 1)$ . For $λ = 1$ , $Q = (5, 6, 1)$ .
Execute: Compute centroid $(α, β, γ)$ of $Δ P Q R$ with vertices $P (- 3, 4, - 1)$ , $Q (5, 6, 1)$ , $R (1, 2, 3)$ : $α = \frac{- 3 + 5 + 1}{3} = 1$ $β = \frac{4 + 6 + 2}{3} = 4$ $γ = \frac{- 1 + 1 + 3}{3} = 1$ Now calculate $α^{2} + β^{2} + γ^{2} = 1^{2} + 4^{2} + 1^{2} = 1 + 16 + 1 = 18$ .
Verify: Verify distance: $P R^{2} = (- 3 - 1)^{2} + (4 - 2)^{2} + (- 1 - 3)^{2} = 16 + 4 + 16 = 36 ⟹ P R = 6$ . $Q R^{2} = (5 - 1)^{2} + (6 - 2)^{2} + (1 - 3)^{2} = 16 + 16 + 4 = 36 ⟹ Q R = 6$ . Both points are verified correctly.

Centroid of a Triangle Formed by Line Points Equidistant from a Given Point

What feels right?

The centroid of triangle $P Q R$ is $(α, β, γ) = (1, 4, 1)$ , giving $α^{2} + β^{2} + γ^{2} = 1^{2} + 4^{2} + 1^{2} = 18$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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