---
title: "Coplanar lines and plane containing them"
subject: "Mathematics"
chapter: 3
key: "semester_2/mathematics:coplanar-lines-plane-containing-them-2"
source: https://www.bctengineer.com/q/semester_2/mathematics/coplanar-lines-plane-containing-them-2
site: BCT Engineering
---

# Coplanar lines and plane containing them

Mathematics · Chapter 3: Three Dimensional Solid Geometry

- Asked in: 2068 Magh, 5 marks
- Video: [Show that the lines `(x-3)/2=(y+1)/(-3)=(z+2)/1 and (x-7)/(-3)=y/1=(z+7)/2` are coplanar. Also find](https://www.youtube.com/watch?v=hQk75QbNNdc) — Doubtnut
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/coplanar-lines-plane-containing-them-2
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-3

## Question

Show that the lines (x+5)/3 = (y+4)/1 = (z−7)/(−2) and 3x+2y+z−2 = 0 = x−3y+2z−13 are coplanar and find the equation of the plane in which they lie.

## Short answer

The plane through the second line and the point (-5,-4,7) on the first is 21x - 19y + 22z - 125 = 0. The direction (3,1,-2) is orthogonal to the normal (21,-19,22), so line 1 lies in the plane; thus the lines are coplanar and this is the containing plane.

## Answer

Method: put the second line as the intersection of two planes, take the plane of the pencil through it which contains a point of the first line, and check that the first line's direction lies in that plane.

Line 1: (x+5)/3 = (y+4)/1 = (z−7)/(−2): point A(−5, −4, 7), direction (3, 1, −2).
Line 2: 3x + 2y + z − 2 = 0 = x − 3y + 2z − 13.

#### Plane through line 2 and the point A
Any plane through line 2:

(3x + 2y + z − 2) + k(x − 3y + 2z − 13) = 0.

Put A(−5, −4, 7):

```text
(−15 − 8 + 7 − 2) + k(−5 + 12 + 14 − 13) = 0
−18 + 8k = 0   →   k = 9/4
```
Plane: 4(3x + 2y + z − 2) + 9(x − 3y + 2z − 13) = 0, i.e.

```text
21x − 19y + 22z − 125 = 0
```

#### The first line lies in this plane
The plane contains A. Its normal is (21, −19, 22) and the direction of line 1 is (3, 1, −2):

21(3) − 19(1) + 22(−2) = 63 − 19 − 44 = 0.

So the direction is perpendicular to the normal; line 1 is parallel to the plane and passes through a point of it, hence it lies in the plane. Both lines are in the same plane, so they are coplanar.

**Answer: the lines are coplanar, and the plane containing them is 21x − 19y + 22z − 125 = 0.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Planes through line 2 form the pencil:
   A. (3x+2y+z-2)(x-3y+2z-13) = 0
   B. 3x+2y+z-2 = k
   C. (3x+2y+z-2) + k(x-3y+2z-13) = 0
   D. k(3x+2y+z-2) = 0
   Answer: C. A combination of the two defining planes. (Key point: Plane through line 2)

2. How is k found?
   A. Make the plane pass through A(-5, -4, 7) on line 1
   B. Make the plane pass through the origin
   C. Make the normal equal (3, 1, -2)
   D. Set k = 1
   Answer: A. This gives -18 + 8k = 0, k = 9/4. (Key point: Choosing k)

3. How is it confirmed that line 1 lies in the plane?
   A. The normal is parallel to direction
   B. A lies on it and the direction (3, 1, -2) is perpendicular to the normal
   C. The plane passes through the origin
   D. k is positive
   Answer: B. Dot product 63 - 19 - 44 = 0. (Key point: Verifying line 1)

4. The plane is:
   A. 21x + 19y + 22z - 125 = 0
   B. 12x - 19y + 22z + 125 = 0
   C. 21x - 19y + 22z + 125 = 0
   D. 21x - 19y + 22z - 125 = 0
   Answer: D. 4(3x+2y+z-2) + 9(x-3y+2z-13) = 21x - 19y + 22z - 125. (Key point: Plane)

## Close variants set in other exams

- [Coplanar lines and plane containing them](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Acoplanar-lines-plane-containing-them) — Mathematics, chapter 3, asked 2×
- [Coplanar lines, intersection point and plane](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Acoplanar-lines-intersection-point-plane) — Mathematics, chapter 3, asked 2×

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