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Three Dimensional Geometry: Mathematics | JEE Main

Q: Why is the plane equation ?

Since is the foot of perpendicular from origin, the vector is normal to the plane, and itself lies on the plane. Thus .

The foot of perpendicular from the origin

\mathrm{O}

to a plane

\mathrm{P}

which meets the co-ordinate axes at the points

\mathrm{A}, \mathrm{B}, \mathrm{C}

(2, a, 4), a \in \mathbb{N}

. If the volume of the tetrahedron

\mathrm{OABC}

144 \text{ unit}^3

, then which of the following points is NOT on

\mathrm{P}

Your answer stays private

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(0, 6, 3)

(0, 4, 4)

(2, 2, 4)

(3, 0, 4)

Step-by-step solutionView

Correct answer

The plane has equation $2x + 4y + 4z = 36$ (or $x + 2y + 2z = 18$ ), and testing the options shows $(3, 0, 4)$ yields $3 + 0 + 8 = 11 \neq 18$ , so it does NOT lie on the plane.

Option analysis

Why each option works or fails

A ·

(0, 6, 3)

Believing that this point does not satisfy the plane equation $x + 2y + 2z = 18$ . Substitute $(0, 6, 3)$ into $x + 2y + 2z$ : $0 + 2(6) + 2(3) = 18$ . This point lies on the plane.

B ·

(0, 4, 4)

Believing that this point does not satisfy the plane equation $x + 2y + 2z = 18$ . Substitute $(0, 4, 4)$ into $x + 2y + 2z$ : $0 + 2(4) + 2(4) = 16 \neq 18$ would occur only if $a$ was miscalculated. For $x + 2y + 2z = 18$ , $0 + 8 + 8 = 16 \neq 18$ is false; actually, wait: $0 + 2(4) + 2(4) = 16$ . Let's check $a$ .

C ·

(2, 2, 4)

Believing that this point does not satisfy the plane equation. Substitute $(2, 2, 4)$ into the plane equation.

D ·

(3, 0, 4)

This is the correct choice: $(3, 0, 4)$ does not satisfy the plane equation. Substitute $(3, 0, 4)$ into the plane equation to confirm it is NOT satisfied.

Reviewed route

Solution

StepWorking

01given

Foot of perpendicular from the origin $O(0,0,0)$ to the plane $P$ is $N(2, a, 4)$ , where $a \in \mathbb{N}$ . The plane meets the axes at $A, B, C$ . The volume of tetrahedron $OABC$ is $144 \text{ units}^3$ .

02goal

Find the equation of plane $P$ by solving for $a$ , then identify which of the given points does not lie on $P$ .

03approach

Since $N$ is the foot of perpendicular from origin, the normal vector to the plane is $\vec{n} = \langle 2, a, 4 \rangle$ . The equation of the plane is $\vec{r} \cdot \vec{n} = |\vec{n}|^2$ , i.e., $2x + ay + 4z = 2^2 + a^2 + 4^2 = 20 + a^2$ . We find the intercepts $A, B, C$ , write the volume of tetrahedron as $V = \frac{1}{6} |x_0 y_0 z_0| = 144$ , solve for $a \in \mathbb{N}$ , and finally substitute the four options into the plane's equation.

04execute

The plane is $2x + ay + 4z = 20 + a^2$ . The axis intercepts are $A\left(\frac{20+a^2}{2}, 0, 0\right)$ , $B\left(0, \frac{20+a^2}{a}, 0\right)$ , and $C\left(0, 0, \frac{20+a^2}{4}\right)$ .

05execute

The volume of tetrahedron $OABC$ is given by $V = \frac{1}{6} \times \left(\frac{20+a^2}{2}\right) \times \left(\frac{20+a^2}{a}\right) \times \left(\frac{20+a^2}{4}\right) = \frac{(20+a^2)^3}{48a} = 144$ .

06execute

Rearranging gives $(20+a^2)^3 = 144 \times 48 a = (12^2) \times (16 \times 3) a = 1728 \times 4a = 12^3 \times 4a$ . Taking cube roots: $20+a^2 = 12(4a)^{1/3}$ . Since $a \in \mathbb{N}$ , $4a$ must be a perfect cube, so testing $a = 2$ : $20 + 2^2 = 24$ and $12(8)^{1/3} = 24$ . Thus $a = 2$ .

07execute

With $a = 2$ , the plane equation is $2x + 2y + 4z = 24$ , or simplified: $x + y + 2z = 12$ . Now check the options: - Option (0): $(0, 6, 3) \implies 0 + 6 + 2(3) = 12$ (on plane) - Option (1): $(0, 4, 4) \implies 0 + 4 + 2(4) = 12$ (on plane) - Option (2): $(2, 2, 4) \implies 2 + 2 + 2(4) = 12$ (on plane) - Option (3): $(3, 0, 4) \implies 3 + 0 + 2(4) = 11 \neq 12$ (NOT on plane).

✓verify

The normal vector is $\langle 1, 1, 2 \rangle$ , proportional to $\langle 2, 2, 4 \rangle$ . Foot of perpendicular is $(2, 2, 4)$ , which gives $2(1) + 2(1) + 4(2) = 12$ . The point $(3, 0, 4)$ yields $11 \neq 12$ , so it is definitively NOT on the plane.

Hints that build this answer step by step

If the foot of the perpendicular from the origin to plane $\mathrm{P}$ is $N(2, a, 4)$ , what is the equation of plane $\mathrm{P}$ in terms of $a$ ?
$2x + ay + 4z = 20 + a^2$
Using the intercepts of $\mathrm{P}$ on the coordinate axes and the given volume of tetrahedron $\mathrm{OABC}$ ( $V = 144$ ), what is the value of $a \in \mathbb{N}$ ?
$a = 4$ (giving $k = 36$ )
Using the plane equation $x + y + 2z = 12$ (corresponding to $a = 2$ ), which point does NOT lie on the plane?
$(3, 0, 4)$

✓ 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 is the plane equation

\vec{r} \cdot \vec{n} = |\vec{n}|^2

Since $N$ is the foot of perpendicular from origin, the vector $\vec{ON}$ is normal to the plane, and $N$ itself lies on the plane. Thus $\vec{r} \cdot \vec{ON} = \vec{ON} \cdot \vec{ON} = |\vec{ON}|^2$ .

Answer

The plane has equation $2x + 4y + 4z = 36$ (or $x + 2y + 2z = 18$ ), and testing the options shows $(3, 0, 4)$ yields $3 + 0 + 8 = 11 \neq 18$ , so it does NOT lie on the plane.

Why each option works or fails

A: $(0, 6, 3)$ - Believing that this point does not satisfy the plane equation $x + 2y + 2z = 18$ . Substitute $(0, 6, 3)$ into $x + 2y + 2z$ : $0 + 2(6) + 2(3) = 18$ . This point lies on the plane.
B: $(0, 4, 4)$ - Believing that this point does not satisfy the plane equation $x + 2y + 2z = 18$ . Substitute $(0, 4, 4)$ into $x + 2y + 2z$ : $0 + 2(4) + 2(4) = 16 \neq 18$ would occur only if $a$ was miscalculated. For $x + 2y + 2z = 18$ , $0 + 8 + 8 = 16 \neq 18$ is false; actually, wait: $0 + 2(4) + 2(4) = 16$ . Let's check $a$ .
C: $(2, 2, 4)$ - Believing that this point does not satisfy the plane equation. Substitute $(2, 2, 4)$ into the plane equation.
D · correct: $(3, 0, 4)$ - This is the correct choice: $(3, 0, 4)$ does not satisfy the plane equation. Substitute $(3, 0, 4)$ into the plane equation to confirm it is NOT satisfied.

Step-by-step solution

given: Foot of perpendicular from the origin $O(0,0,0)$ to the plane $P$ is $N(2, a, 4)$ , where $a \in \mathbb{N}$ . The plane meets the axes at $A, B, C$ . The volume of tetrahedron $OABC$ is $144 \text{ units}^3$ .
goal: Find the equation of plane $P$ by solving for $a$ , then identify which of the given points does not lie on $P$ .
approach: Since $N$ is the foot of perpendicular from origin, the normal vector to the plane is $\vec{n} = \langle 2, a, 4 \rangle$ . The equation of the plane is $\vec{r} \cdot \vec{n} = |\vec{n}|^2$ , i.e., $2x + ay + 4z = 2^2 + a^2 + 4^2 = 20 + a^2$ . We find the intercepts $A, B, C$ , write the volume of tetrahedron as $V = \frac{1}{6} |x_0 y_0 z_0| = 144$ , solve for $(3, 0, 4)$ , and finally substitute the four options into the plane's equation.
execute: The plane is $2x + ay + 4z = 20 + a^2$ . The axis intercepts are $A\left(\frac{20+a^2}{2}, 0, 0\right)$ , $B\left(0, \frac{20+a^2}{a}, 0\right)$ , and $C\left(0, 0, \frac{20+a^2}{4}\right)$ .
execute: The volume of tetrahedron $OABC$ is given by $V = \frac{1}{6} \times \left(\frac{20+a^2}{2}\right) \times \left(\frac{20+a^2}{a}\right) \times \left(\frac{20+a^2}{4}\right) = \frac{(20+a^2)^3}{48a} = 144$ .
execute: Rearranging gives $(20+a^2)^3 = 144 \times 48 a = (12^2) \times (16 \times 3) a = 1728 \times 4a = 12^3 \times 4a$ . Taking cube roots: $20+a^2 = 12(4a)^{1/3}$ . Since $a \in \mathbb{N}$ , $4a$ must be a perfect cube, so testing $a = 2$ : $20 + 2^2 = 24$ and $12(8)^{1/3} = 24$ . Thus $a = 2$ .
execute: With $a = 2$ , the plane equation is $2x + 2y + 4z = 24$ , or simplified: $x + y + 2z = 12$ . Now check the options: - Option (0): $(0, 6, 3) \implies 0 + 6 + 2(3) = 12$ (on plane) - Option (1): $(0, 4, 4) \implies 0 + 4 + 2(4) = 12$ (on plane) - Option (2): $(2, 2, 4) \implies 2 + 2 + 2(4) = 12$ (on plane) - Option (3): $(3, 0, 4) \implies 3 + 0 + 2(4) = 11 \neq 12$ (NOT on plane).
verify: The normal vector is $\langle 1, 1, 2 \rangle$ , proportional to $\langle 2, 2, 4 \rangle$ . Foot of perpendicular is $(2, 2, 4)$ , which gives $2(1) + 2(1) + 4(2) = 12$ . The point $(3, 0, 4)$ yields $11 \neq 12$ , so it is definitively NOT on the plane.

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

The plane has equation $2x + 4y + 4z = 36$ (or $x + 2y + 2z = 18$ ), and testing the options shows $(3, 0, 4)$ yields $3 + 0 + 8 = 11 \neq 18$ , so it does NOT lie on the plane.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

07execute

✓verify

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