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Three Dimensional Geometry: Mathematics | JEE Main

Let the plane

\mathrm{P} : 8x + \alpha_1 y + \alpha_2 z + 12 = 0

be parallel to the line

\mathrm{L} : \frac{x+2}{2} = \frac{y-3}{3} = \frac{z+4}{5}

. If the intercept of

\mathrm{P}

on the

y

-axis is

1

, then the distance between

\mathrm{P}

and

\mathrm{L}

is :

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.

\sqrt{\frac{6}{7}}

\sqrt{\frac{7}{2}}

\frac{6}{\sqrt{14}}

\sqrt{14}

Step-by-step solutionView

Correct answer

The plane is $8x - 12y + 4z + 12 = 0$ (or $2x - 3y + z + 3 = 0$ ), and the perpendicular distance from the point $(-2, 3, -4)$ on the line to this plane is $\sqrt{14}$ .

Option analysis

Why each option works or fails

A ·

\sqrt{\frac{6}{7}}

Computing distance using an unreduced normal vector but incorrectly evaluating the numerator or taking the square root of the ratio. Ensure the formula for the perpendicular distance $d = \frac{|a x_0 + b y_0 + c z_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$ is evaluated directly without introducing extraneous square roots over the numerator.

B ·

\sqrt{\frac{7}{2}}

Dividing out factors inconsistently between the numerator and the denominator inside the distance formula. Simplify the equation of the plane completely to $2x - 3y + z + 3 = 0$ first, then compute $|2(-2) - 3(3) + 1(-4) + 3| / \sqrt{2^2 + (-3)^2 + 1^2} = 14 / \sqrt{14} = \sqrt{14}$ .

C ·

\frac{6}{\sqrt{14}}

Miscalculating the numerator $|2(-2) - 3(3) + 1(-4) + 3|$ as $6$ instead of $14$ by dropping or misapplying signs on the coordinates. Carefully substitute the point on the line $(-2, 3, -4)$ : $2(-2) - 3(3) + 1(-4) + 3 = -4 - 9 - 4 + 3 = -14$ , giving absolute value $14$ .

D ·

\sqrt{14}

None. This is the correct option. The values $\alpha_1 = -12$ and $\alpha_2 = 4$ give the plane $2x - 3y + z + 3 = 0$ , which is at distance $\frac{|-14|}{\sqrt{14}} = \sqrt{14}$ from $(-2, 3, -4)$ .

Reviewed route

Solution

StepWorking

01given

Plane $P: 8x + \alpha_1 y + \alpha_2 z + 12 = 0$ , line $L: \frac{x+2}{2} = \frac{y-3}{3} = \frac{z+4}{5}$ , $y$ -intercept of $P$ is $1$ , and $P \parallel L$ .

02goal

Determine the values of $\alpha_1$ and $\alpha_2$ , and calculate the perpendicular distance between plane $P$ and line $L$ .

03approach

First, find $\alpha_1$ using the condition that $(0, 1, 0)$ lies on $P$ . Next, use the condition that $P \parallel L \implies \vec{n} \cdot \vec{d}_L = 0$ to find $\alpha_2$ . Simplify the plane's equation and find the perpendicular distance from a known point on the line, $(-2, 3, -4)$ , to plane $P$ .

04execute

Substitute $(0, 1, 0)$ into $8(0) + \alpha_1(1) + \alpha_2(0) + 12 = 0 \implies \alpha_1 = -12$ .

05execute

Normal to the plane is $\vec{n} = (8, \alpha_1, \alpha_2) = (8, -12, \alpha_2)$ and line direction is $\vec{d} = (2, 3, 5)$ . Since $P \parallel L$ , $\vec{n} \cdot \vec{d} = 0 \implies 8(2) - 12(3) + 5\alpha_2 = 0 \implies 16 - 36 + 5\alpha_2 = 0 \implies 5\alpha_2 = 20 \implies \alpha_2 = 4$ .

06execute

The equation of the plane is $8x - 12y + 4z + 12 = 0$ , which simplifies on dividing by 4 to $2x - 3y + z + 3 = 0$ . A point on line $L$ is $(-2, 3, -4)$ . Distance $d = \frac{|2(-2) - 3(3) + 1(-4) + 3|}{\sqrt{2^2 + (-3)^2 + 1^2}} = \frac{|-4 - 9 - 4 + 3|}{\sqrt{4 + 9 + 1}} = \frac{14}{\sqrt{14}} = \sqrt{14}$ .

✓verify

Check that the direction ratios $(2, -3, 1)$ are orthogonal to line $(2, 3, 5)$ : $2(2) - 3(3) + 1(5) = 4 - 9 + 5 = 0$ . The distance evaluates to $\sqrt{14}$ , matching option (3).

Hints that build this answer step by step

Given that the $y$ -intercept of the plane $\mathrm{P}: 8x + \alpha_1 y + \alpha_2 z + 12 = 0$ is $1$ , what is the value of $\alpha_1$ ?
$\alpha_1 = -12$
The plane $\mathrm{P}$ with normal $\mathbf{n} = \langle 8, -12, \alpha_2 \rangle$ is parallel to the line $\mathrm{L}$ with direction vector $\mathbf{d} = \langle 2, 3, 5 \rangle$ . What is the condition relating $\mathbf{n}$ and $\mathbf{d}$ , and what is $\alpha_2$ ?
$\mathbf{n} \cdot \mathbf{d} = 0$ , which yields $\alpha_2 = 4$
What is the perpendicular distance from the point $(-2, 3, -4)$ on line $\mathrm{L}$ to the plane $2x - 3y + z + 3 = 0$ ?
$\sqrt{14}$

✓ 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

Can we pick any point on the line

L

to find the distance to plane

P

Yes, because the line is parallel to the plane, every point on the line is at the same perpendicular distance from the plane.

Answer

The plane is $8x - 12y + 4z + 12 = 0$ (or $2x - 3y + z + 3 = 0$ ), and the perpendicular distance from the point $(-2, 3, -4)$ on the line to this plane is $\sqrt{14}$ .

Why each option works or fails

A: $\sqrt{\frac{6}{7}}$ - Computing distance using an unreduced normal vector but incorrectly evaluating the numerator or taking the square root of the ratio. Ensure the formula for the perpendicular distance $d = \frac{|a x_0 + b y_0 + c z_0 + d|}{\sqrt{a^2 + b^2 + c^2}}$ is evaluated directly without introducing extraneous square roots over the numerator.
B: $\sqrt{\frac{7}{2}}$ - Dividing out factors inconsistently between the numerator and the denominator inside the distance formula. Simplify the equation of the plane completely to $2x - 3y + z + 3 = 0$ first, then compute $|2(-2) - 3(3) + 1(-4) + 3| / \sqrt{2^2 + (-3)^2 + 1^2} = 14 / \sqrt{14} = \sqrt{14}$ .
C: $\frac{6}{\sqrt{14}}$ - Miscalculating the numerator $|2(-2) - 3(3) + 1(-4) + 3|$ as $6$ instead of $14$ by dropping or misapplying signs on the coordinates. Carefully substitute the point on the line $(-2, 3, -4)$ : $2(-2) - 3(3) + 1(-4) + 3 = -4 - 9 - 4 + 3 = -14$ , giving absolute value $14$ .
D · correct: $\sqrt{14}$ - None. This is the correct option. The values $\alpha_1 = -12$ and $\alpha_2 = 4$ give the plane $2x - 3y + z + 3 = 0$ , which is at distance $\frac{|-14|}{\sqrt{14}} = \sqrt{14}$ from $(-2, 3, -4)$ .

Step-by-step solution

given: Plane $P: 8x + \alpha_1 y + \alpha_2 z + 12 = 0$ , line $L: \frac{x+2}{2} = \frac{y-3}{3} = \frac{z+4}{5}$ , $y$ -intercept of $P$ is $1$ , and $P \parallel L$ .
goal: Determine the values of $\alpha_1$ and $\alpha_2$ , and calculate the perpendicular distance between plane $P$ and line $L$ .
approach: First, find $\alpha_1$ using the condition that $(0, 1, 0)$ lies on $P$ . Next, use the condition that $P \parallel L \implies \vec{n} \cdot \vec{d}_L = 0$ to find $\alpha_2$ . Simplify the plane's equation and find the perpendicular distance from a known point on the line, $(-2, 3, -4)$ , to plane $P$ .
execute: Substitute $(0, 1, 0)$ into $8(0) + \alpha_1(1) + \alpha_2(0) + 12 = 0 \implies \alpha_1 = -12$ .
execute: Normal to the plane is $\vec{n} = (8, \alpha_1, \alpha_2) = (8, -12, \alpha_2)$ and line direction is $\vec{d} = (2, 3, 5)$ . Since $P \parallel L$ , $\vec{n} \cdot \vec{d} = 0 \implies 8(2) - 12(3) + 5\alpha_2 = 0 \implies 16 - 36 + 5\alpha_2 = 0 \implies 5\alpha_2 = 20 \implies \alpha_2 = 4$ .
execute: The equation of the plane is $8x - 12y + 4z + 12 = 0$ , which simplifies on dividing by 4 to $2x - 3y + z + 3 = 0$ . A point on line $L$ is $(-2, 3, -4)$ . Distance $d = \frac{|2(-2) - 3(3) + 1(-4) + 3|}{\sqrt{2^2 + (-3)^2 + 1^2}} = \frac{|-4 - 9 - 4 + 3|}{\sqrt{4 + 9 + 1}} = \frac{14}{\sqrt{14}} = \sqrt{14}$ .
verify: Check that the direction ratios $(2, -3, 1)$ are orthogonal to line $(2, 3, 5)$ : $2(2) - 3(3) + 1(5) = 4 - 9 + 5 = 0$ . The distance evaluates to $\sqrt{14}$ , matching option (3).

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

The plane is $8x - 12y + 4z + 12 = 0$ (or $2x - 3y + z + 3 = 0$ ), and the perpendicular distance from the point $(-2, 3, -4)$ on the line to this plane is $\sqrt{14}$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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