JEE Main 2026 (22 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Three Dimensional Geometry: Let $L$ be the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 3}{6}$ and let $S$ be the set of all points $(a, b, c)$ on $L$ , whose distance from the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z - 9}{0}$ along the line $L$ is 7. Then $\sum_{(a, b, c) \in S} (a + b + c)$ is equal to :

Q: Is there a shortcut to finding the sum of coordinates without calculating the points individually?

You could potentially use the properties of the parameter $\lambda$. Since the coordinates are linear functions of $\lambda$, the sum $a+b+c$ is also a linear function of $\lambda$. If you find the sum for the intersection point and then apply the change in $\lambda$, you might save a few seconds, but calculating the points is safer and less prone to conceptual errors!

Select Answer:

Progressive Hints

Hint 1

Hint 1: Visualize the two lines in 3D space. The phrase 'distance along the line L' is the key. It implies that we are not looking for the shortest perpendicular distance between two skew lines, but rather the distance from a specific reference point on line L to any other point on the same line. Can you identify the point where these two lines meet?

Hint 2

Hint 2: Once you have identified the intersection point

M

, realize that any point on line

L

can be expressed in terms of a single parameter

λ

. Use the distance formula between your general point

P (λ)

and the fixed point

M

to set up an algebraic equation.

Hint 3

Hint 3: The distance condition is given as 7. Set the distance

P M = 7

. When you square both sides, you will find a quadratic in

λ

. The equation will look like

49 (λ - 2)^{2} = 49

. Solve for

λ

and find the two corresponding points.

Visualized Solution

Visualizing the Problem

Line $L_{1}$ : $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 3}{6}$
Line $L_{2}$ : $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z - 9}{0}$
Distance along the line $L_{1}$ from $L_{2}$ means the distance from their intersection point.

The Intersection Point $M$

Let the intersection point of $L_{1}$ and $L_{2}$ be $M$ .
We need to find the coordinates of $M$ first.

Parameterizing Line $L_{1}$

Let $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 3}{6} = λ$
General point on $L_{1}$ : $P (λ) = (2 λ - 1, 3 λ - 1, 6 λ - 3)$

Analyzing Line $L_{2}$

Line $L_{2}$ : $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z - 9}{0}$
The denominator of the $z$ -term is $0$ .
This implies $z - 9 = 0 ⟹ z = 9$ for all points on $L_{2}$ .

Finding Intersection Parameter $λ$

At intersection $M$ , the $z$ -coordinate of $L_{1}$ must equal the $z$ -coordinate of $L_{2}$ .
$6 λ - 3 = 9$
$6 λ = 12 ⟹ λ = 2$

Coordinates of Intersection Point $M$

Substitute $λ = 2$ into the general point of $L_{1}$ :
$x = 2 (2) - 1 = 3$
$y = 3 (2) - 1 = 5$
$z = 6 (2) - 3 = 9$
Intersection Point $M = (3, 5, 9)$

Setting Up the Distance Condition

Let $P (2 λ - 1, 3 λ - 1, 6 λ - 3)$ be a point in set $S$ .
Distance $P M = 7$
$(2 λ - 1 - 3)^{2} + (3 λ - 1 - 5)^{2} + (6 λ - 3 - 9)^{2} = 7$

Simplifying the Distance Equation

Simplify inside the square root:
$(2 λ - 4)^{2} + (3 λ - 6)^{2} + (6 λ - 12)^{2} = 7$
Factor out common terms:
$4 (λ - 2)^{2} + 9 (λ - 2)^{2} + 36 (λ - 2)^{2} = 7$

Solving for Parameter $λ$

Combine the terms:
$49 (λ - 2)^{2} = 7$
Take the square root:
$7∣ λ - 2∣ = 7 ⟹ ∣ λ - 2∣ = 1$

Finding the Two Values of $λ$

Remove the absolute value:
$λ - 2 = 1 ⟹ λ = 3$
$λ - 2 = - 1 ⟹ λ = 1$

Coordinates and Sum for $λ = 3$

For $λ = 3$ , substitute into the general point:
Point $P = (2 (3) - 1, 3 (3) - 1, 6 (3) - 3) = (5, 8, 15)$
Sum of coordinates $a + b + c = 5 + 8 + 15 = 28$

Coordinates and Sum for $λ = 1$

For $λ = 1$ , substitute into the general point:
Point $Q = (2 (1) - 1, 3 (1) - 1, 6 (1) - 3) = (1, 2, 3)$
Sum of coordinates $a + b + c = 1 + 2 + 3 = 6$

Final Calculation

The set $S$ contains points $P$ and $Q$ .
Total sum $= \sum_{(a, b, c) \in S} (a + b + c)$
Total sum $= 28 + 6 = 34$
The final answer is 34.

00:00 / 00:00

The Sigma Insight: Equation of a Line in Space

Solution Diagram

Analyzing the Setup

Imagine you are standing in a vast, three-dimensional coordinate space. You see two lines,

L_{1}

and

L_{2}

, stretching out into the void.

The line

L_{1}

is defined by the equation:

\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 3}{6}

You are traveling along this path, and you need to find specific locations that are exactly

7

units away from a reference point defined by the second line,

L_{2}

The Intersection

Finding Our Anchor

Before we can measure distance, we need a starting point. The problem implies that the two lines share a common point, which we will call

M

The second line is defined as:

\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z - 9}{0}

The zero in the denominator indicates that the

z

-coordinate is locked at

z = 9

. By setting the

z

-coordinate of a general point on

L_{1}

, given by

P (λ) = (2 λ - 1, 3 λ - 1, 6 λ - 3)

, equal to

9

, we find:

6 λ - 3 = 9 \Rightarrow 6 λ = 12 \Rightarrow λ = 2

Substituting

λ = 2

back into the coordinates of

P

, we discover our anchor point

M = (3, 5, 9)

. This point serves as the origin for our distance measurements.

The Dance of the Parameter

We seek points

P

L_{1}

such that the distance

P M = 7

. We represent any point on

L_{1}

P (2 λ - 1, 3 λ - 1, 6 λ - 3)

Using the distance formula, we set up the following equation:

(2 λ - 1 - 3)^{2} + (3 λ - 1 - 5)^{2} + (6 λ - 3 - 9)^{2} = 7

Simplifying the terms inside the square root, we obtain:

(2 λ - 4)^{2} + (3 λ - 6)^{2} + (6 λ - 12)^{2} = 7

Factoring out the common term

(λ - 2)

, we get:

4 (λ - 2)^{2} + 9 (λ - 2)^{2} + 36 (λ - 2)^{2} = 7

Summing the coefficients, we find

4 + 9 + 36 = 49

. Since

49 = 7

, the equation simplifies to:

7∣ λ - 2∣ = 7 \Rightarrow ∣ λ - 2∣ = 1

The Final Destination

The absolute value equation

∣ λ - 2∣ = 1

yields two possible values for the parameter: 1.

λ - 2 = 1 \Rightarrow λ = 3

λ - 2 = - 1 \Rightarrow λ = 1

For

λ = 3

, the point

P

is:

P = (2 (3) - 1, 3 (3) - 1, 6 (3) - 3) = (5, 8, 15)

The sum of these coordinates is

5 + 8 + 15 = 28

For

λ = 1

, the point

Q

is:

Q = (2 (1) - 1, 3 (1) - 1, 6 (1) - 3) = (1, 2, 3)

The sum of these coordinates is

1 + 2 + 3 = 6

The sum of all such coordinates for all points in the set is

28 + 6 = 34

. The final answer is 34.

Common Traps & FAQs

Wait, why do we need to find the intersection point

M

? Can't we just pick any point on

L_{2}

Great question! The problem specifies the distance 'along the line

L

'. This implies that the reference point for our distance measurement is the point where

L

and

L_{2}

intersect. If you pick an arbitrary point on

L_{2}

, you aren't measuring distance 'along

L

'. By finding the intersection, we anchor our coordinate system to the unique point that connects both lines.

The denominator of the

z

-term in

L_{2}

is 0. Does that mean the line is undefined?

Not at all! In 3D geometry, a denominator of 0 in the symmetric form

\frac{z - z _{0}}{c} = 0

simply means the line is parallel to the

x y

-plane. It tells us that for every point on that line, the

z

-coordinate is constant. In this case,

z = 9

for all points on

L_{2}

Why do we get two values for

λ

? Does that mean there are two points?

Exactly! Think of it like a number line. If you are at a point

M

and you want to find all points at a distance of 7 units from you, you can go 7 units in the 'positive' direction along the line or 7 units in the 'negative' direction. Both points satisfy the condition, which is why we get two distinct values for

λ

Is there a shortcut to finding the sum of coordinates without calculating the points individually?

You could potentially use the properties of the parameter

λ

. Since the coordinates are linear functions of

λ

, the sum

a + b + c

is also a linear function of

λ

. If you find the sum for the intersection point and then apply the change in

λ

, you might save a few seconds, but calculating the points is safer and less prone to conceptual errors!

What if the lines were skew and didn't intersect? Would the problem still make sense?

If the lines were skew, the phrase 'distance along the line

L

from

L_{2}

' would be ambiguous because there would be no single point of intersection. The problem relies on the existence of a common point to define the 'zero' distance. In JEE problems, always check for intersection first!

Select Answer:

Visualized Solution

Visualizing the Problem

The Intersection Point $M$

Parameterizing Line $L_{1}$

Analyzing Line $L_{2}$

Finding Intersection Parameter $λ$

Coordinates of Intersection Point $M$

Setting Up the Distance Condition

Simplifying the Distance Equation

Solving for Parameter $λ$

Finding the Two Values of $λ$

Coordinates and Sum for $λ = 3$

Coordinates and Sum for $λ = 1$

Final Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Intersection

The Dance of the Parameter

The Final Destination

Similar Questions

If the distances of the point $(1, 2, a)$ from the line $\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1}$ along the lines $L_{1} : \frac{x - 1}{3} = \frac{y - 2}{4} = \frac{z - a}{b}$ and $L_{2} : \frac{x - 1}{1} = \frac{y - 2}{4} = \frac{z - a}{c}$ are equal, then $a + b + c$ is equal to

Let $d$ be the distance of the point of intersection of the lines $\frac{x + 6}{3} = \frac{y}{2} = \frac{z + 1}{1}$ and $\frac{x - 7}{4} = \frac{y - 9}{3} = \frac{z - 4}{2}$ from the point $(7, 8, 9)$ . Then $d^{2} + 6$ is equal to :

Consider the line $L$ passing through the points $(1, 2, 3)$ and $(2, 3, 5)$ . The distance of the point $(\frac{11}{3}, \frac{11}{3}, \frac{19}{3})$ from the line $L$ along the line $\frac{3 x - 11}{2} = \frac{3 y - 11}{1} = \frac{3 z - 19}{2}$ is equal to

The distance, of the point $(7, - 2, 11)$ from the line $\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}$ along the line $\frac{x - 5}{2} = \frac{y - 1}{- 3} = \frac{z - 5}{6}$ , is :

The distance of the point $(7, 10, 11)$ from the line $\frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3}$ along the line $\frac{x - 9}{2} = \frac{y - 13}{3} = \frac{z - 17}{6}$ is

If the length of the perpendicular drawn from the point $P (a, 4, 2), a > 0$ on the line $\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1}$ is $26$ units and $Q (α_{1}, α_{2}, α_{3})$ is the image of the point $P$ in this line, then $a + \sum_{i = 1}^{3} α_{i}$ is equal to :

The square of the distance of the point $(\frac{15}{7}, \frac{32}{7}, 7)$ from the line $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ in the direction of the vector $\hat{i} + 4 \hat{j} + 7 \hat{k}$ is:

Let the point, on the line passing through the points $P (1, - 2, 3)$ and $Q (5, - 4, 7)$ , farther from the origin and at distance of 9 units from the point $P$ , be $(α, β, γ)$ . Then $α^{2} + β^{2} + γ^{2}$ is equal to :

Let in a $△ A B C$ the length of the side AC be 6, the vertex B be (1, 2, 3) and the vertices A, C lie on the line $\frac{x-6}{3}= rac{y-7}{2}= rac{z-7}{-2}.$ Then the area (in sq. units) of $△ A B C$ is:

Select Answer:

Visualized Solution

Visualizing the Problem

The Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Parameterizing Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L1​

Analyzing Line .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } L2​

Finding Intersection Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Coordinates of Intersection Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Setting Up the Distance Condition

Simplifying the Distance Equation

Solving for Parameter .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Finding the Two Values of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Coordinates and Sum for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ=3

Coordinates and Sum for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ=1

Final Calculation

The Sigma Insight: Equation of a Line in Space

Analyzing the Setup

The Intersection

The Dance of the Parameter

The Final Destination

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let d be the distance of the point of intersection of the lines 3x+6​=2y​=1z+1​ and 4x−7​=3y−9​=2z−4​ from the point (7,8,9). Then d2+6 is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance, of the point (7,−2,11) from the line 1x−6​=0y−4​=3z−8​ along the line 2x−5​=−3y−1​=6z−5​, is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The distance of the point (7,10,11) from the line 1x−4​=0y−4​=3z−2​ along the line 2x−9​=3y−13​=6z−17​ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The square of the distance of the point (715​,732​,7) from the line 3x+1​=5y+3​=7z+5​ in the direction of the vector i^+4j^​+7k^ is:

The Intersection Point $M$

Parameterizing Line $L_{1}$

Analyzing Line $L_{2}$

Finding Intersection Parameter $λ$

Coordinates of Intersection Point $M$

Solving for Parameter $λ$

Finding the Two Values of $λ$

Coordinates and Sum for $λ = 3$

Coordinates and Sum for $λ = 1$

Let $d$ be the distance of the point of intersection of the lines $\frac{x + 6}{3} = \frac{y}{2} = \frac{z + 1}{1}$ and $\frac{x - 7}{4} = \frac{y - 9}{3} = \frac{z - 4}{2}$ from the point $(7, 8, 9)$ . Then $d^{2} + 6$ is equal to :

The distance, of the point $(7, - 2, 11)$ from the line $\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}$ along the line $\frac{x - 5}{2} = \frac{y - 1}{- 3} = \frac{z - 5}{6}$ , is :

The distance of the point $(7, 10, 11)$ from the line $\frac{x - 4}{1} = \frac{y - 4}{0} = \frac{z - 2}{3}$ along the line $\frac{x - 9}{2} = \frac{y - 13}{3} = \frac{z - 17}{6}$ is

The square of the distance of the point $(\frac{15}{7}, \frac{32}{7}, 7)$ from the line $\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}$ in the direction of the vector $\hat{i} + 4 \hat{j} + 7 \hat{k}$ is: