JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Image of a point in a plane in three-dimensional geometry

Let (α, β, γ) be the image of the point P(2,3,5) in the plane 2x + y - 3z = 6. Then α + β + γ is equal to

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A10B5C12D9

Step-by-step solutionView

Correct answer

The image of the point $(2, 3, 5)$ in the plane $2 x + y - 3 z - 6 = 0$ is $(6, 5, - 1)$ , giving $α + β + γ = 10$ .

Option analysis

Why each option works or fails

A · 10

This is the correct solution obtained by applying the reflection formula for a point in a plane. Correctly compute the scale factor $k = \frac{- 2 (2 (2) + 3 - 3 (5) - 6)}{2^{2} + 1^{2} + (- 3)^{2}} = \frac{- 2 (- 14)}{14} = 2$ , which gives $(α, β, γ) = (6, 5, - 1)$ and sum $10$ .

B · 5

The student computed the coordinates of the foot of the perpendicular instead of the reflection (image), omitting the factor of 2. To find the image rather than the foot of the perpendicular, multiply the projection ratio by $- 2$ instead of $- 1$ .

C · 12

The student made a sign error when substituting the point into the plane equation, taking $2 (2) + 3 - 3 (5) + 6$ instead of $- 6$ . Write the plane in the standard form $a x + b y + c z + d = 0$ ; here $d = - 6$ , not $+ 6$ .

D · 9

The student made an arithmetic error when adding the components $α + β + γ = 6 + 5 + (- 1)$ or in calculating the denominator $a^{2} + b^{2} + c^{2}$ . Double-check the arithmetic: $6 + 5 - 1 = 10$ , not $9$ .

Reviewed route

Solution

StepWorking

01Given

Point $P (x_{1}, y_{1}, z_{1}) = (2, 3, 5)$ and plane equation $a x + b y + c z + d = 0$ is $2 x + y - 3 z - 6 = 0$ .

02Goal

Find the image $(α, β, γ)$ of point $P$ and compute the sum $α + β + γ$ .

03Approach

Use the standard reflection formula in 3D geometry: $\frac{α - x_{1}}{a} = \frac{β - y_{1}}{b} = \frac{γ - z_{1}}{c} = \frac{- 2 (a x_{1} + b y_{1} + c z_{1} + d)}{a^{2} + b^{2} + c^{2}}$ .

04Execute

Compute the ratio factor $k = \frac{- 2 (2 (2) + 1 (3) - 3 (5) - 6)}{2^{2} + 1^{2} + (- 3)^{2}} = \frac{- 2 (4 + 3 - 15 - 6)}{4 + 1 + 9} = \frac{- 2 (- 14)}{14} = 2$ .

05Execute

Find the coordinates: $α = 2 + 2 (2) = 6$ $β = 3 + 1 (2) = 5$ $γ = 5 - 3 (2) = - 1$ Sum: $α + β + γ = 6 + 5 - 1 = 10$ .

✓Verify

The midpoint of $P (2, 3, 5)$ and $(α, β, γ) = (6, 5, - 1)$ is $M (\frac{2 + 6}{2}, \frac{3 + 5}{2}, \frac{5 - 1}{2}) = (4, 4, 2)$ . Check if $M$ lies on the plane: $2 (4) + 4 - 3 (2) = 8 + 4 - 6 = 6$ , which holds true.

Hints that build this answer step by step

What formula gives the coordinates of the image $(α, β, γ)$ of a point $(x_{1}, y_{1}, z_{1})$ in the plane $a x + b y + c z + d = 0$ ?
$\frac{α - x_{1}}{a} = \frac{β - y_{1}}{b} = \frac{γ - z_{1}}{c} = \frac{- 2 (a x_{1} + b y_{1} + c z_{1} + d)}{a^{2} + b^{2} + c^{2}}$
Evaluating the ratio $k = \frac{- 2 (2 (2) + 1 (3) - 3 (5) - 6)}{2^{2} + 1^{2} + (- 3)^{2}}$ , what is the value of $k$ ?
$k = 2$
Using $k = 2$ with $α = 2 + 2 a$ , $β = 3 + 2 b$ , and $γ = 5 + 2 c$ , what is the image point $(α, β, γ)$ ?
$(6, 5, - 1)$
What is the final sum $α + β + γ$ ?
$10$

✓ 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 is the factor multiplied by -2 rather than -1?

Multiplying by -1 gives the foot of the perpendicular (the midpoint between the point and its image). Since the image is twice as far along the normal vector, we use -2.

Answer

The image of the point $(2, 3, 5)$ in the plane $2 x + y - 3 z - 6 = 0$ is $(6, 5, - 1)$ , giving $α + β + γ = 10$ .

Why each option works or fails

A · correct: 10 - This is the correct solution obtained by applying the reflection formula for a point in a plane. Correctly compute the scale factor $k = \frac{- 2 (2 (2) + 3 - 3 (5) - 6)}{2^{2} + 1^{2} + (- 3)^{2}} = \frac{- 2 (- 14)}{14} = 2$ , which gives $(α, β, γ) = (6, 5, - 1)$ and sum $10$ .
B: 5 - The student computed the coordinates of the foot of the perpendicular instead of the reflection (image), omitting the factor of 2. To find the image rather than the foot of the perpendicular, multiply the projection ratio by $- 2$ instead of $- 1$ .
C: 12 - The student made a sign error when substituting the point into the plane equation, taking $2 (2) + 3 - 3 (5) + 6$ instead of $- 6$ . Write the plane in the standard form $a x + b y + c z + d = 0$ ; here $d = - 6$ , not $+ 6$ .
D: 9 - The student made an arithmetic error when adding the components $α + β + γ = 6 + 5 + (- 1)$ or in calculating the denominator $a^{2} + b^{2} + c^{2}$ . Double-check the arithmetic: $6 + 5 - 1 = 10$ , not $9$ .

Step-by-step solution

Given: Point $P (x_{1}, y_{1}, z_{1}) = (2, 3, 5)$ and plane equation $a x + b y + c z + d = 0$ is $2 x + y - 3 z - 6 = 0$ .
Goal: Find the image $(α, β, γ)$ of point $P$ and compute the sum $α + β + γ$ .
Approach: Use the standard reflection formula in 3D geometry: $\frac{α - x_{1}}{a} = \frac{β - y_{1}}{b} = \frac{γ - z_{1}}{c} = \frac{- 2 (a x_{1} + b y_{1} + c z_{1} + d)}{a^{2} + b^{2} + c^{2}}$ .
Execute: Compute the ratio factor $k = \frac{- 2 (2 (2) + 1 (3) - 3 (5) - 6)}{2^{2} + 1^{2} + (- 3)^{2}} = \frac{- 2 (4 + 3 - 15 - 6)}{4 + 1 + 9} = \frac{- 2 (- 14)}{14} = 2$ .
Execute: Find the coordinates: $α = 2 + 2 (2) = 6$ $6 + 5 - 1 = 10$ $γ = 5 - 3 (2) = - 1$ Sum: $α + β + γ = 6 + 5 - 1 = 10$ .
Verify: The midpoint of $P (2, 3, 5)$ and $(α, β, γ) = (6, 5, - 1)$ is $M (\frac{2 + 6}{2}, \frac{3 + 5}{2}, \frac{5 - 1}{2}) = (4, 4, 2)$ . Check if $M$ lies on the plane: $2 (4) + 4 - 3 (2) = 8 + 4 - 6 = 6$ , which holds true.

Image of a point in a plane in three-dimensional geometry

What feels right?

The image of the point $(2, 3, 5)$ in the plane $2 x + y - 3 z - 6 = 0$ is $(6, 5, - 1)$ , giving $α + β + γ = 10$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

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