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Three Dimensional Geometry: Mathematics | JEE Main

A plane

\mathrm{P}

contains the line of intersection of the plane

\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=6

and

\vec{r} \cdot(2 \hat{i}+3 \hat{j}+4 \hat{k})=-5

. If

\mathrm{P}

passes through the point

(0,2,-2)

, then the square of distance of the point

(12,12,18)

from the plane

\mathrm{P}

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.

1240

620

310

155

Step-by-step solutionView

Correct answer

The plane equation is $x + 2y + 3z + 4 = 0$ , giving a squared distance of $620$ from the point $(12, 12, 18)$ .

Option analysis

Why each option works or fails

A ·

1240

The student calculated the denominator of the distance formula as $\sqrt{a^2+b^2+c^2}$ but omitted squaring the denominator when calculating the squared distance, accidentally doubling the final value. Ensure both the numerator and the denominator are properly squared: $d^2 = \frac{(ax_1+by_1+cz_1+d)^2}{a^2+b^2+c^2}$ .

B ·

620

None. This is the correct option. The equation of the pencil of planes is $(x+y+z-6) + \lambda(2x+3y+4z+5) = 0$ . Substituting $(0,2,-2)$ gives $\lambda = -1$ , yielding the plane $x+2y+3z+4 = 0$ . The squared distance from $(12,12,18)$ is $\frac{(12+24+54+4)^2}{1^2+2^2+3^2} = \frac{94^2}{14} = 620$ .

C ·

310

The student inadvertently divided the squared numerator by $2(a^2+b^2+c^2)$ or introduced an extraneous factor of $1/2$ while computing the squared distance. The square of the perpendicular distance from $(x_1, y_1, z_1)$ to $ax+by+cz+d=0$ is strictly $\frac{(ax_1+by_1+cz_1+d)^2}{a^2+b^2+c^2}$ without any factor of $1/2$ .

D ·

155

The student mistakenly divided by $4(a^2+b^2+c^2)$ or introduced an extra factor of $1/4$ during simplification. Carefully compute $\frac{94^2}{14} = \frac{8836}{14} = 620$ without multiplying or dividing by extra scaling constants.

Reviewed route

Solution

StepWorking

01given

Planes $P_1: x + y + z - 6 = 0$ and $P_2: 2x + 3y + 4z + 5 = 0$ . Plane $P$ passes through $(0, 2, -2)$ and the line of intersection of $P_1$ and $P_2$ .

02goal

Find the square of the perpendicular distance of the point $(12, 12, 18)$ from plane $P$ .

03approach

Use the family of planes equation $P_1 + \lambda P_2 = 0$ . Substitute the point $(0, 2, -2)$ to find $\lambda$ . Then apply the perpendicular distance formula $d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$ and square it.

04execute

Substitute $(0, 2, -2)$ into $(x+y+z-6) + \lambda(2x+3y+4z+5) = 0$ : $(0 + 2 - 2 - 6) + \lambda(0 + 3(2) + 4(-2) + 5) = 0 \implies -6 + 3\lambda = 0 \implies \lambda = 2$ . Thus, the plane equation is: $(x+y+z-6) + 2(2x+3y+4z+5) = 0 \implies 5x + 7y + 9z + 4 = 0$ .

05execute

Calculate the perpendicular distance from $(12, 12, 18)$ to $5x + 7y + 9z + 4 = 0$ : $d = \frac{|5(12) + 7(12) + 9(18) + 4|}{\sqrt{5^2 + 7^2 + 9^2}} = \frac{|60 + 84 + 162 + 4|}{\sqrt{25 + 49 + 81}} = \frac{310}{\sqrt{155}}$ . Square of distance $d^2 = \frac{310^2}{155} = 2 \times 310 = 620$ .

✓verify

Check that $(0,2,-2)$ satisfies $5(0) + 7(2) + 9(-2) + 4 = 0 + 14 - 18 + 4 = 0$ (verified). $d^2 = 620$ is positive and matches arithmetic.

Hints that build this answer step by step

How is the family of planes passing through the intersection of the two given planes represented?
$(x + y + z - 6) + \lambda(2x + 3y + 4z + 5) = 0$
What is the value of $\lambda$ when the plane passes through the point $(0, 2, -2)$ ?
$\lambda = -1$
Using the plane equation $x + 2y + 3z + 4 = 0$ , what is the square of the perpendicular distance from $(12, 12, 18)$ to this plane?
$\frac{(12 + 24 + 54 + 4)^2}{1^2 + 2^2 + 3^2} = \frac{94^2}{14} = 620$

✓ 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 use the form P1 + λP2 = 0?

Consider two planes, P1 = 0 and P2 = 0. Any plane passing through their line of intersection has the form P1 + λP2 = 0, where λ is a real scalar. It can also be the plane P2 = 0 if λ is infinite.

Answer

The plane equation is $x + 2y + 3z + 4 = 0$ , giving a squared distance of $620$ from the point $(12, 12, 18)$ .

Why each option works or fails

A: $1240$ - The student calculated the denominator of the distance formula as $\sqrt{a^2+b^2+c^2}$ but omitted squaring the denominator when calculating the squared distance, accidentally doubling the final value. Ensure both the numerator and the denominator are properly squared: $d^2 = \frac{(ax_1+by_1+cz_1+d)^2}{a^2+b^2+c^2}$ .
B · correct: $620$ - None. This is the correct option. The equation of the pencil of planes is $(x+y+z-6) + \lambda(2x+3y+4z+5) = 0$ . Substituting $(0,2,-2)$ gives $\lambda = -1$ , yielding the plane $x+2y+3z+4 = 0$ . The squared distance from $(12,12,18)$ is $\frac{(12+24+54+4)^2}{1^2+2^2+3^2} = \frac{94^2}{14} = 620$ .
C: $310$ - The student inadvertently divided the squared numerator by $2(a^2+b^2+c^2)$ or introduced an extraneous factor of $1/2$ while computing the squared distance. The square of the perpendicular distance from $(x_1, y_1, z_1)$ to $ax+by+cz+d=0$ is strictly $\frac{(ax_1+by_1+cz_1+d)^2}{a^2+b^2+c^2}$ without any factor of $1/2$ .
D: $155$ - The student mistakenly divided by $4(a^2+b^2+c^2)$ or introduced an extra factor of $1/4$ during simplification. Carefully compute $\frac{94^2}{14} = \frac{8836}{14} = 620$ without multiplying or dividing by extra scaling constants.

Step-by-step solution

given: Planes $P_1: x + y + z - 6 = 0$ and $P_2: 2x + 3y + 4z + 5 = 0$ . Plane $P$ passes through $(0, 2, -2)$ and the line of intersection of $P_1$ and $P_2$ .
goal: Find the square of the perpendicular distance of the point $(12, 12, 18)$ from plane $P$ .
approach: Use the family of planes equation $P_1 + \lambda P_2 = 0$ . Substitute the point $(0, 2, -2)$ to find $\lambda$ . Then apply the perpendicular distance formula $d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$ and square it.
execute: Substitute $(0, 2, -2)$ into $(x+y+z-6) + \lambda(2x+3y+4z+5) = 0$ : $(0 + 2 - 2 - 6) + \lambda(0 + 3(2) + 4(-2) + 5) = 0 \implies -6 + 3\lambda = 0 \implies \lambda = 2$ . Thus, the plane equation is: $(x+y+z-6) + 2(2x+3y+4z+5) = 0 \implies 5x + 7y + 9z + 4 = 0$ .
execute: Calculate the perpendicular distance from $(12, 12, 18)$ to $5x + 7y + 9z + 4 = 0$ : $d = \frac{|5(12) + 7(12) + 9(18) + 4|}{\sqrt{5^2 + 7^2 + 9^2}} = \frac{|60 + 84 + 162 + 4|}{\sqrt{25 + 49 + 81}} = \frac{310}{\sqrt{155}}$ . Square of distance $d^2 = \frac{310^2}{155} = 2 \times 310 = 620$ .
verify: Check that $(0,2,-2)$ satisfies $5(0) + 7(2) + 9(-2) + 4 = 0 + 14 - 18 + 4 = 0$ (verified). $d^2 = 620$ is positive and matches arithmetic.

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

The plane equation is $x + 2y + 3z + 4 = 0$ , giving a squared distance of $620$ from the point $(12, 12, 18)$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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