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Three Dimensional Geometry: Mathematics | JEE Main

The plane, passing through the points

(0, -1, 2)

and

(-1, 2, 1)

and parallel to the line passing through

(5, 1, -7)

and

(1, -1, -1)

, also passes through the point.

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.

(1, -2, 1)

(0, 5, -2)

(-2, 5, 0)

(2, 0, 1)

Step-by-step solutionView

Correct answer

The plane has normal vector $(1, 2, 5)$ and equation $x + 2y + 5z = 8$ , which contains the point $(-2, 5, 0)$ .

Option analysis

Why each option works or fails

A ·

(1, -2, 1)

Substituting $(1, -2, 1)$ into $x + 2y + 5z$ yields $1 - 4 + 5 = 2$ , mistakenly equating it to the constant or miscalculating the sum as 8. Evaluate the plane equation $x + 2y + 5z$ carefully: $1 + 2(-2) + 5(1) = 2 \neq 8$ .

B ·

(0, 5, -2)

Evaluating $(0, 5, -2)$ gives $0 + 10 - 10 = 0$ , mistakenly thinking the plane passes through the origin or confusing the RHS constant. Check the value against the constant $d = 8$ : $0 + 2(5) + 5(-2) = 0 \neq 8$ .

C ·

(-2, 5, 0)

None. This point satisfies the equation of the plane: $-2 + 2(5) + 5(0) = 8$ . This is the correct option.

D ·

(2, 0, 1)

Substituting $(2, 0, 1)$ gives $2 + 0 + 5 = 7$ , confusing 7 with 8 due to an arithmetic slip. Evaluate carefully: $2 + 2(0) + 5(1) = 7 \neq 8$ , so the point does not lie on the plane.

Reviewed route

Solution

StepWorking

01given

The plane passes through $A(0, -1, 2)$ and $B(-1, 2, 1)$ , and is parallel to the line joining $C(5, 1, -7)$ and $D(1, -1, -1)$ .

02goal

Find which of the four given points lies on the plane.

03approach

Find two vectors parallel to the plane: $\vec{v}_1 = \vec{AB}$ and $\vec{v}_2 = \vec{CD}$ . The normal vector to the plane is $\vec{n} = \vec{v}_2 \times \vec{v}_1$ . Then write the equation of the plane $\vec{n} \cdot (\vec{r} - \vec{r}_A) = 0$ and substitute each option.

04execute

Vector in the plane: $\vec{AB} = (-1 - 0)\hat{i} + (2 - (-1))\hat{j} + (1 - 2)\hat{k} = -\hat{i} + 3\hat{j} - \hat{k}$ . Vector parallel to the line: $\vec{CD} = (5 - 1)\hat{i} + (1 - (-1))\hat{j} + (-7 - (-1))\hat{k} = 4\hat{i} + 2\hat{j} - 6\hat{k}$ . Normal vector $\vec{n} = \vec{CD} \times \vec{AB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 2 & -6 \\ -1 & 3 & -1 \end{vmatrix} = \hat{i}(-2 + 18) - \hat{j}(-4 - 6) + \hat{k}(12 + 2) = 16\hat{i} + 10\hat{j} + 14\hat{k}$ . Dividing by 2 gives the simplified normal vector: $\vec{n}' = 8\hat{i} + 5\hat{j} + 7\hat{k}$ .

05execute

Equation of the plane is $8x + 5y + 7z = d$ . Substituting $A(0, -1, 2)$ : $d = 8(0) + 5(-1) + 7(2) = -5 + 14 = 9$ . So, the plane equation is $8x + 5y + 7z = 9$ .

06execute

Test the given options against $8x + 5y + 7z = 9$ : (0) $(1, -2, 1) \implies 8(1) + 5(-2) + 7(1) = 8 - 10 + 7 = 5 \neq 9$ (1) $(0, 5, -2) \implies 8(0) + 5(5) + 7(-2) = 25 - 14 = 11 \neq 9$ (2) $(-2, 5, 0) \implies 8(-2) + 5(5) + 7(0) = -16 + 25 + 0 = 9$ (Matches!) (3) $(2, 0, 1) \implies 8(2) + 5(0) + 7(1) = 16 + 7 = 23 \neq 9$

✓verify

Check point $B(-1, 2, 1)$ on the plane: $8(-1) + 5(2) + 7(1) = -8 + 10 + 7 = 9$ , which holds. Check dot product with line vector $(2, 1, -3)$ : $8(2) + 5(1) + 7(-3) = 16 + 5 - 21 = 0$ , which is perpendicular to normal. Hence, the plane equation is robustly verified.

Hints that build this answer step by step

What are two non-parallel vectors parallel to the required plane?
Vector between the two points: $(-1, 3, -1)$ ; direction vector of the line: $(-2, -1, 3)$
What is the normal vector $\vec{n}$ to the plane found by taking the cross product $(-1, 3, -1) \times (-2, -1, 3)$ ?
$(1, 2, 5)$ or any scalar multiple like $(8, 5, 7)$ ? Specifically, $(8, 5, 7)$ is incorrect; the cross product is $(8, 5, 7)$ ? No, $(3)(3) - (-1)(-1) = 8$ ; $-((-1)(3) - (-1)(-2)) = 5$ ; $(-1)(-1) - (3)(-2) = 7$ , wait, let's recompute: $(8, 5, 7)$ . Let's check: $8(0) + 5(-1) + 7(2) = 9$ . Then $-2(8)+5(5)+0 = 9$ .
Using the normal vector $(8, 5, 7)$ and the point $(0, -1, 2)$ , the plane equation is $8x + 5y + 7z = 9$ . Which given point satisfies this equation?
$(-2, 5, 0)$

✓ 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 we take the cross product of the two direction vectors?

The vector joining the two points on the plane and the line's direction vector are both parallel to the plane. Therefore, their cross product is orthogonal to the plane, which gives its normal vector.

Answer

The plane has normal vector $(1, 2, 5)$ and equation $x + 2y + 5z = 8$ , which contains the point $(-2, 5, 0)$ .

Why each option works or fails

A: $(1, -2, 1)$ - Substituting $(1, -2, 1)$ into $x + 2y + 5z$ yields $1 - 4 + 5 = 2$ , mistakenly equating it to the constant or miscalculating the sum as 8. Evaluate the plane equation $x + 2y + 5z$ carefully: $1 + 2(-2) + 5(1) = 2 \neq 8$ .
B: $(0, 5, -2)$ - Evaluating $(0, 5, -2)$ gives $0 + 10 - 10 = 0$ , mistakenly thinking the plane passes through the origin or confusing the RHS constant. Check the value against the constant $d = 8$ : $0 + 2(5) + 5(-2) = 0 \neq 8$ .
C · correct: $(-2, 5, 0)$ - None. This point satisfies the equation of the plane: $-2 + 2(5) + 5(0) = 8$ . This is the correct option.
D: $(2, 0, 1)$ - Substituting $(2, 0, 1)$ gives $2 + 0 + 5 = 7$ , confusing 7 with 8 due to an arithmetic slip. Evaluate carefully: $2 + 2(0) + 5(1) = 7 \neq 8$ , so the point does not lie on the plane.

Step-by-step solution

given: The plane passes through $A(0, -1, 2)$ and $1 + 2(-2) + 5(1) = 2 \neq 8$ , and is parallel to the line joining $C(5, 1, -7)$ and $D(1, -1, -1)$ .
goal: Find which of the four given points lies on the plane.
approach: Find two vectors parallel to the plane: $\vec{v}_1 = \vec{AB}$ and $\vec{v}_2 = \vec{CD}$ . The normal vector to the plane is $\vec{n} = \vec{v}_2 \times \vec{v}_1$ . Then write the equation of the plane $\vec{n} \cdot (\vec{r} - \vec{r}_A) = 0$ and substitute each option.
execute: Vector in the plane: $\vec{AB} = (-1 - 0)\hat{i} + (2 - (-1))\hat{j} + (1 - 2)\hat{k} = -\hat{i} + 3\hat{j} - \hat{k}$ . Vector parallel to the line: $\vec{CD} = (5 - 1)\hat{i} + (1 - (-1))\hat{j} + (-7 - (-1))\hat{k} = 4\hat{i} + 2\hat{j} - 6\hat{k}$ . Normal vector $\vec{n} = \vec{CD} \times \vec{AB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 2 & -6 \\ -1 & 3 & -1 \end{vmatrix} = \hat{i}(-2 + 18) - \hat{j}(-4 - 6) + \hat{k}(12 + 2) = 16\hat{i} + 10\hat{j} + 14\hat{k}$ . Dividing by 2 gives the simplified normal vector: $\vec{n}' = 8\hat{i} + 5\hat{j} + 7\hat{k}$ .
execute: Equation of the plane is $8x + 5y + 7z = d$ . Substituting $A(0, -1, 2)$ : $d = 8(0) + 5(-1) + 7(2) = -5 + 14 = 9$ . So, the plane equation is $8x + 5y + 7z = 9$ .
execute: Test the given options against $8x + 5y + 7z = 9$ : (0) $(1, -2, 1) \implies 8(1) + 5(-2) + 7(1) = 8 - 10 + 7 = 5 \neq 9$ (1) $(0, 5, -2) \implies 8(0) + 5(5) + 7(-2) = 25 - 14 = 11 \neq 9$ (2) $(-2, 5, 0) \implies 8(-2) + 5(5) + 7(0) = -16 + 25 + 0 = 9$ (Matches!) (3) $(2, 0, 1) \implies 8(2) + 5(0) + 7(1) = 16 + 7 = 23 \neq 9$
verify: Check point $B(-1, 2, 1)$ on the plane: $8(-1) + 5(2) + 7(1) = -8 + 10 + 7 = 9$ , which holds. Check dot product with line vector $(2, 1, -3)$ : $8(2) + 5(1) + 7(-3) = 16 + 5 - 21 = 0$ , which is perpendicular to normal. Hence, the plane equation is robustly verified.

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

The plane has normal vector $(1, 2, 5)$ and equation $x + 2y + 5z = 8$ , which contains the point $(-2, 5, 0)$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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