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Three Dimensional Geometry: Mathematics | JEE Main

If the equation of the plane passing through the point

(1,1,2)

and perpendicular to the line

x - 3y + 2z - 1 = 0 = 4x - y + z

Ax + By + Cz = 1

, then

140(C - B + A)

is equal to

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Step-by-step solutionView

Correct answer

The value of 140(C - B + A) is -28.

Option analysis

Why each option works or fails

Reviewed route

Solution

StepWorking

01given

A plane passes through $(1, 1, 2)$ and is perpendicular to the line given by $x - 3y + 2z - 1 = 0 = 4x - y + z$ . The equation of the plane is $Ax + By + Cz = 1$ .

02goal

Find the value of $140(C - B + A)$ .

03approach

The line is given as the intersection of two planes with normals $\vec{n}_1 = \langle 1, -3, 2 \rangle$ and $\vec{n}_2 = \langle 4, -1, 1 \rangle$ . The direction vector $\vec{d}$ of the line is along $\vec{n}_1 \times \vec{n}_2$ . Since the required plane is perpendicular to the line, its normal vector $\vec{n}$ is parallel to $\vec{d}$ . With $\vec{n}$ and point $(1, 1, 2)$ , write the plane's equation, normalize it so that the constant on the right-hand side is $1$ , and determine $A, B, C$ .

04execute

Compute the cross product: $\vec{n} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 4 & -1 & 1 \end{vmatrix} = \hat{i}((-3)(1) - (2)(-1)) - \hat{j}((1)(1) - (2)(4)) + \hat{k}((1)(-1) - (-3)(4)) = -\hat{i} + 7\hat{j} + 11\hat{k}$ .

05execute

The plane equation through $(1,1,2)$ with normal $\langle -1, 7, 11 \rangle$ is $-1(x - 1) + 7(y - 1) + 11(z - 2) = 0 \implies -x + 7y + 11z = -1 + 7 + 22 = 28$ .

06execute

Divide by $28$ to match $Ax + By + Cz = 1$ : $-\frac{1}{28}x + \frac{7}{28}y + \frac{11}{28}z = 1$ , giving $A = -\frac{1}{28}$ , $B = \frac{7}{28}$ , $C = \frac{11}{28}$ .

07execute

Compute $140(C - B + A) = 140\left(\frac{11 - 7 - 1}{28}\right) = 140 \times \frac{3}{28} = 5 \times 3 = 15$ .

✓verify

Check that $(1, 1, 2)$ lies on $-x + 7y + 11z = 28$ : $-(1) + 7(1) + 11(2) = -1 + 7 + 22 = 28$ , which holds. Also, $\langle -1, 7, 11 \rangle \cdot \langle 1, -3, 2 \rangle = -1 - 21 + 22 = 0$ and $\langle -1, 7, 11 \rangle \cdot \langle 4, -1, 1 \rangle = -4 - 7 + 11 = 0$ , confirming normality.

Hints that build this answer step by step

How can the normal vector to the plane be determined from the given line equations?
Calculate the cross product of the normal vectors of the two planes defining the line.
What is the cross product of the normals n₁ = (1, -3, 2) and n₂ = (4, -1, 1)?
(-1, 7, 11)
Using the normal vector (-1, 7, 11) and the point (1, 1, 2), what is the equation of the plane in the form Ax + By + Cz = 1?
(-1/28)x + (7/28)y + (11/28)z = 1
What is the value of 140(C - B + A) using A = -1/28, B = 7/28, and C = 11/28?
15

✓ Source and academic review

Question type: Numerical
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 is the direction vector of the line given by the cross product of the normal vectors?

The line is the intersection of two planes, so it lies within both planes simultaneously. Thus, its direction must be perpendicular to the normal vectors of both planes.

Answer

The value of 140(C - B + A) is -28.

Why each option works or fails

Step-by-step solution

given: A plane passes through $(1, 1, 2)$ and is perpendicular to the line given by $x - 3y + 2z - 1 = 0 = 4x - y + z$ . The equation of the plane is $Ax + By + Cz = 1$ .
goal: Find the value of $140(C - B + A)$ .
approach: The line is given as the intersection of two planes with normals $\vec{n}_1 = \langle 1, -3, 2 \rangle$ and $\vec{n}_2 = \langle 4, -1, 1 \rangle$ . The direction vector $\vec{d}$ of the line is along $\vec{n}_1 \times \vec{n}_2$ . Since the required plane is perpendicular to the line, its normal vector $\vec{n}$ is parallel to $\vec{d}$ . With $\vec{n}$ and point $(1, 1, 2)$ , write the plane's equation, normalize it so that the constant on the right-hand side is $1$ , and determine $A, B, C$ .
execute: Compute the cross product: $\vec{n} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 4 & -1 & 1 \end{vmatrix} = \hat{i}((-3)(1) - (2)(-1)) - \hat{j}((1)(1) - (2)(4)) + \hat{k}((1)(-1) - (-3)(4)) = -\hat{i} + 7\hat{j} + 11\hat{k}$ .
execute: The plane equation through $(1,1,2)$ with normal $\langle -1, 7, 11 \rangle$ is $-1(x - 1) + 7(y - 1) + 11(z - 2) = 0 \implies -x + 7y + 11z = -1 + 7 + 22 = 28$ .
execute: Divide by $28$ to match $Ax + By + Cz = 1$ : $-\frac{1}{28}x + \frac{7}{28}y + \frac{11}{28}z = 1$ , giving $A = -\frac{1}{28}$ , $B = \frac{7}{28}$ , $C = \frac{11}{28}$ .
execute: Compute $140(C - B + A) = 140\left(\frac{11 - 7 - 1}{28}\right) = 140 \times \frac{3}{28} = 5 \times 3 = 15$ .
verify: Check that $(1, 1, 2)$ lies on $-x + 7y + 11z = 28$ : $-(1) + 7(1) + 11(2) = -1 + 7 + 22 = 28$ , which holds. Also, $\langle -1, 7, 11 \rangle \cdot \langle 1, -3, 2 \rangle = -1 - 21 + 22 = 0$ and $\langle -1, 7, 11 \rangle \cdot \langle 4, -1, 1 \rangle = -4 - 7 + 11 = 0$ , confirming normality.

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

The value of 140(C - B + A) is -28.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

07execute

✓verify

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