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Three Dimensional Geometry: Mathematics | JEE Main

Let

\mathrm{P}

and

\mathrm{Q}

be the points on the line

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

which are at a distance of

6

units from the point

\mathrm{R}(1,2,3)

. If the centroid of the triangle

\mathrm{PQR}

(\alpha, \beta, \gamma)

, then

\alpha^2 + \beta^2 + \gamma^2

is:

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

26

36

18

24

Step-by-step solutionView

Correct answer

The centroid of triangle $\mathrm{PQR}$ is $(\alpha, \beta, \gamma) = (1, 4, 1)$ , giving $\alpha^2 + \beta^2 + \gamma^2 = 1^2 + 4^2 + 1^2 = 18$ .

Option analysis

Why each option works or fails

A ·

26

Calculated the sum of squares of the midpoint of $\mathrm{PQ}$ instead of the centroid of triangle $\mathrm{PQR}$ . Remember that the centroid coordinates of $\triangle\mathrm{PQR}$ involve averaging the coordinates of all three points $\mathrm{P}$ , $\mathrm{Q}$ , and $\mathrm{R}$ , not just the midpoint of $\mathrm{PQ}$ .

B ·

36

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

C ·

18

None. The solution correctly identifies the coordinates of $\mathrm{P}$ and $\mathrm{Q}$ , evaluates the centroid of $\triangle\mathrm{PQR}$ , and sums the squares of its components. None.

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: $\alpha = 1, \beta = 4, \gamma = 1 \implies 1^2 + 4^2 + 1^2 = 1 + 16 + 1 = 18$ .

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 $PR = QR = 6$ .

02goal

Find the centroid $(\alpha, \beta, \gamma)$ of $\Delta PQR$ and compute $\alpha^2 + \beta^2 + \gamma^2$ .

03approach

Express any point on the line in terms of parameter $\lambda$ as $(8\lambda - 3, 2\lambda + 4, 2\lambda - 1)$ . 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 $\lambda$ , find coordinates of $P$ and $Q$ , compute the centroid of $\Delta PQR$ , and evaluate $\alpha^2 + \beta^2 + \gamma^2$ .

04execute

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

05execute

Compute centroid $(\alpha, \beta, \gamma)$ of $\Delta PQR$ with vertices $P(-3, 4, -1)$ , $Q(5, 6, 1)$ , $R(1, 2, 3)$ : $\alpha = \frac{-3 + 5 + 1}{3} = 1$ $\beta = \frac{4 + 6 + 2}{3} = 4$ $\gamma = \frac{-1 + 1 + 3}{3} = 1$ Now calculate $\alpha^2 + \beta^2 + \gamma^2 = 1^2 + 4^2 + 1^2 = 1 + 16 + 1 = 18$ .

✓verify

Verify distance: $PR^2 = (-3-1)^2 + (4-2)^2 + (-1-3)^2 = 16 + 4 + 16 = 36 \implies PR = 6$ . $QR^2 = (5-1)^2 + (6-2)^2 + (1-3)^2 = 16 + 16 + 4 = 36 \implies QR = 6$ . Both points are verified correctly.

Hints that build this answer step by step

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

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
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 $\lambda$ must yield precisely the two points.

Answer

The centroid of triangle $\mathrm{PQR}$ is $(\alpha, \beta, \gamma) = (1, 4, 1)$ , giving $\alpha^2 + \beta^2 + \gamma^2 = 1^2 + 4^2 + 1^2 = 18$ .

Why each option works or fails

A: $26$ - Calculated the sum of squares of the midpoint of $\mathrm{PQ}$ instead of the centroid of triangle $\mathrm{PQR}$ . Remember that the centroid coordinates of $\triangle\mathrm{PQR}$ involve averaging the coordinates of all three points $\mathrm{P}$ , $\mathrm{Q}$ , and $\mathrm{R}$ , not just the midpoint of $(\alpha, \beta, \gamma)$ .
B: $36$ - Made an algebraic sign error when expanding the distance equation, leading to incorrect parameter values for $\mathrm{P}$ and $\mathrm{Q}$ . Carefully expand $(8\lambda - 4)^2 + (2\lambda + 2)^2 + (2\lambda - 4)^2 = 36$ to find the roots $\lambda = 0, 1$ with the correct signs.
C · correct: $18$ - None. The solution correctly identifies the coordinates of $\mathrm{P}$ and $\mathrm{Q}$ , evaluates the centroid of $\triangle\mathrm{PQR}$ , and sums the squares of its components. None.
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: $\alpha = 1, \beta = 4, \gamma = 1 \implies 1^2 + 4^2 + 1^2 = 1 + 16 + 1 = 18$ .

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 $PR = QR = 6$ .
goal: Find the centroid $(\alpha, \beta, \gamma)$ of $\Delta PQR$ and compute $\alpha^2 + \beta^2 + \gamma^2$ .
approach: Express any point on the line in terms of parameter $\lambda$ as $(8\lambda - 3, 2\lambda + 4, 2\lambda - 1)$ . 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 $\lambda$ , find coordinates of $P$ and $Q$ , compute the centroid of $\Delta PQR$ , and evaluate $\alpha^2 + \beta^2 + \gamma^2$ .
execute: Using the distance formula from $(8\lambda - 3, 2\lambda + 4, 2\lambda - 1)$ to $(1, 2, 3)$ : $(8\lambda - 4)^2 + (2\lambda + 2)^2 + (2\lambda - 4)^2 = 36$ Expanding the terms: $(64\lambda^2 - 64\lambda + 16) + (4\lambda^2 + 8\lambda + 4) + (4\lambda^2 - 16\lambda + 16) = 36$ $72\lambda^2 - 72\lambda + 36 = 36 \implies 72\lambda(\lambda - 1) = 0$ Thus, $\lambda = 0$ or $\lambda = 1$ . For $\lambda = 0$ , $P = (-3, 4, -1)$ . For $\lambda = 1$ , $Q = (5, 6, 1)$ .
execute: Compute centroid $(\alpha, \beta, \gamma)$ of $\Delta PQR$ with vertices $P(-3, 4, -1)$ , $Q(5, 6, 1)$ , $R(1, 2, 3)$ : $\alpha = \frac{-3 + 5 + 1}{3} = 1$ $\beta = \frac{4 + 6 + 2}{3} = 4$ $\gamma = \frac{-1 + 1 + 3}{3} = 1$ Now calculate $\alpha^2 + \beta^2 + \gamma^2 = 1^2 + 4^2 + 1^2 = 1 + 16 + 1 = 18$ .
verify: Verify distance: $PR^2 = (-3-1)^2 + (4-2)^2 + (-1-3)^2 = 16 + 4 + 16 = 36 \implies PR = 6$ . $QR^2 = (5-1)^2 + (6-2)^2 + (1-3)^2 = 16 + 16 + 4 = 36 \implies QR = 6$ . Both points are verified correctly.

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

The centroid of triangle $\mathrm{PQR}$ is $(\alpha, \beta, \gamma) = (1, 4, 1)$ , giving $\alpha^2 + \beta^2 + \gamma^2 = 1^2 + 4^2 + 1^2 = 18$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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Three Dimensional Geometry: Mathematics | JEE Main

The centroid of triangle PQR\mathrm{PQR}PQR is (α,β,γ)=(1,4,1)(\alpha, \beta, \gamma) = (1, 4, 1)(α,β,γ)=(1,4,1), giving α2+β2+γ2=12+42+12=18\alpha^2 + \beta^2 + \gamma^2 = 1^2 + 4^2 + 1^2 = 18α2+β2+γ2=12+42+12=18.

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