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Three Dimensional Geometry: Mathematics | JEE Main

Let the shortest distance between the lines

L : \frac{x-5}{-2} = \frac{y-\lambda}{0} = \frac{z+\lambda}{1}, \lambda \ge 0

and

L_1 : x + 1 = y - 1 = 4 - z

2\sqrt{6}

. If

(\alpha, \beta, \gamma)

lies on

\text{L}

, then which of the following is NOT possible ?

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.

\alpha - 2\gamma = 19

\alpha + \gamma = 7

2\alpha - \gamma = 9

\alpha + 2\gamma = 24

Step-by-step solutionView

Correct answer

Using the shortest distance formula between skew lines yields $\lambda = 7$ , which constrains any point on $L$ such that $\alpha + 2\gamma = 5 - 2\lambda = -9$ , making $\alpha + 2\gamma = 24$ impossible.

Option analysis

Why each option works or fails

A ·

\alpha - 2\gamma = 19

Believing that $\alpha - 2\gamma = 19$ is inconsistent with the parametric coordinates of line $L$ . Express $(\alpha, \beta, \gamma)$ as $(5 - 2k, \lambda, -\lambda + k)$ . The equation $\alpha - 2\gamma = 19$ solves to $(5 - 2k) - 2(k - 7) = 19 \implies -4k + 19 = 19 \implies k = 0$ , which is completely valid.

B · 2

\alpha + \gamma = 7

Assuming $2\alpha + \gamma = 7$ cannot be satisfied by any point on the line $L$ . Substitute the parametric form into the relation: $2(5 - 2k) + (k - 7) = 3 - 3k = 7 \implies k = -4/3$ . Since $k$ can be any real number, this relation is possible.

C ·

2\alpha - \gamma = 9

Assuming $2\alpha - \gamma = 9$ cannot hold for any point on $L$ . Substitute the parametric form: $2(5 - 2k) - (k - 7) = 17 - 5k = 9 \implies k = 8/5$ . This yields a valid point on $L$ , so it is possible.

D ·

\alpha + 2\gamma = 24

Correct choice: recognized that $\alpha + 2\gamma$ eliminates the parameter $k$ entirely and yields a constant value unequal to $24$ . Notice that $\alpha + 2\gamma = (5 - 2k) + 2(k - \lambda) = 5 - 2\lambda$ . With $\lambda = 7$ , $\alpha + 2\gamma = -9$ identically for all points on $L$ , meaning it can never equal $24$ .

Reviewed route

Solution

StepWorking

01given

Line $L: \frac{x-5}{-2} = \frac{y-\lambda}{0} = \frac{z+\lambda}{1}$ with $\lambda \ge 0$ , and line $L_1: \frac{x+1}{1} = \frac{y-1}{1} = \frac{z-4}{-1}$ . Shortest distance $d = 2\sqrt{6}$ . A general point on $L$ is $(\alpha, \beta, \gamma)$ .

02goal

Find the value of $\lambda \ge 0$ using the shortest distance formula, express $(\alpha, \beta, \gamma)$ in terms of a single parameter, and determine which of the given linear relations among $\alpha$ and $\gamma$ is NOT possible.

03approach

Express the shortest distance between skew lines via $d = \frac{|(\vec{a}_1 - \vec{a}_2) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$ . Solve for $\lambda$ using $\lambda \ge 0$ . Then parameterize $(\alpha, \gamma)$ on line $L$ and test the given options.

04execute

Compute $\vec{b}_1 \times \vec{b}_2$ where $\vec{b}_1 = \langle -2, 0, 1 \rangle$ and $\vec{b}_2 = \langle 1, 1, -1 \rangle$ : $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 0 & 1 \\ 1 & 1 & -1 \end{vmatrix} = \hat{i}(0 - 1) - \hat{j}(2 - 1) + \hat{k}(-2 - 0) = \langle -1, -1, -2 \rangle$ Its magnitude is $|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-1)^2 + (-1)^2 + (-2)^2} = \sqrt{6}$ .

05execute

Let $\vec{a}_1 = \langle 5, \lambda, -\lambda \rangle$ and $\vec{a}_2 = \langle -1, 1, 4 \rangle$ . Then $\vec{a}_1 - \vec{a}_2 = \langle 6, \lambda - 1, -\lambda - 4 \rangle$ . Calculate the scalar triple product: $(\vec{a}_1 - \vec{a}_2) \cdot (\vec{b}_1 \times \vec{b}_2) = 6(-1) + (\lambda - 1)(-1) + (-\lambda - 4)(-2) = -6 - \lambda + 1 + 2\lambda + 8 = \lambda + 3$ Setting up the shortest distance equation: $d = \frac{|\lambda + 3|}{\sqrt{6}} = 2\sqrt{6} \implies |\lambda + 3| = 12$ Since $\lambda \ge 0$ , we have $\lambda + 3 = 12 \implies \lambda = 9$ (discarding $\lambda = -15$ ).

06execute

Any point $(\alpha, \beta, \gamma)$ on $L$ satisfies: $\frac{\alpha - 5}{-2} = \frac{\beta - 9}{0} = \frac{\gamma + 9}{1} = k$ So $\alpha = 5 - 2k$ , $\beta = 9$ , $\gamma = -9 + k$ . Notice that $\alpha + 2\gamma = (5 - 2k) + 2(-9 + k) = 5 - 18 = -13$ , which is a constant independent of $k$ . Thus $\alpha + 2\gamma$ MUST always equal $-13$ . Consequently, $\alpha + 2\gamma = 24$ is impossible.

✓verify

Check the other options for consistency: Option (0): $\alpha - 2\gamma = 5 - 2k - 2(-9 + k) = 23 - 4k = 19 \implies k = 1$ (possible). Option (1): $2\alpha + \gamma = 2(5 - 2k) + (-9 + k) = 1 - 3k = 7 \implies k = -2$ (possible). Option (2): $2\alpha - \gamma = 2(5 - 2k) - (-9 + k) = 19 - 5k = 9 \implies k = 2$ (possible). Hence only Option (3) cannot be satisfied.

Hints that build this answer step by step

What are the standard direction vectors $\vec{b}_1, \vec{b}_2$ and points $\vec{a}_1, \vec{a}_2$ on lines $L$ and $L_1$ ?
$\vec{a}_1 = (5, \lambda, -\lambda)$ , $\vec{b}_1 = (-2, 0, 1)$ and $\vec{a}_2 = (-1, 1, 4)$ , $\vec{b}_2 = (1, 1, -1)$
Using the shortest distance formula $d = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|} = 2\sqrt{6}$ , what is the value of $\lambda$ given $\lambda \ge 0$ ?
$\lambda = 7$
Any point on $L$ can be parameterized as $(\alpha, \beta, \gamma) = (5 - 2k, 7, -7 + k)$ . Which linear combination of $\alpha$ and $\gamma$ is independent of the parameter $k$ ?
$\alpha + 2\gamma$ , which identically equals $-9$

✓ 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 does

\alpha + 2\gamma

not depend on

k

Because the direction ratios of the line are proportional to $\langle -2, 0, 1 \rangle$ , any linear combination with coefficients matching the orthogonal relation $1(-2) + 2(1) = 0$ will cancel the parameter $k$ entirely, yielding a constant.

Answer

Using the shortest distance formula between skew lines yields $\lambda = 7$ , which constrains any point on $L$ such that $\alpha + 2\gamma = 5 - 2\lambda = -9$ , making $\alpha + 2\gamma = 24$ impossible.

Why each option works or fails

A: $\alpha - 2\gamma = 19$ - Believing that $\alpha - 2\gamma = 19$ is inconsistent with the parametric coordinates of line $L$ . Express $(\alpha, \beta, \gamma)$ as $(5 - 2k, \lambda, -\lambda + k)$ . The equation $\alpha - 2\gamma = 19$ solves to $(5 - 2k) - 2(k - 7) = 19 \implies -4k + 19 = 19 \implies k = 0$ , which is completely valid.
B: 2 $\alpha + \gamma = 7$ - Assuming $2\alpha + \gamma = 7$ cannot be satisfied by any point on the line $L$ . Substitute the parametric form into the relation: $2(5 - 2k) + (k - 7) = 3 - 3k = 7 \implies k = -4/3$ . Since $k$ can be any real number, this relation is possible.
C: $2\alpha - \gamma = 9$ - Assuming $2\alpha - \gamma = 9$ cannot hold for any point on $L$ . Substitute the parametric form: $2(5 - 2k) - (k - 7) = 17 - 5k = 9 \implies k = 8/5$ . This yields a valid point on $L$ , so it is possible.
D · correct: $\alpha + 2\gamma = 24$ - Correct choice: recognized that $\alpha + 2\gamma$ eliminates the parameter $k$ entirely and yields a constant value unequal to $24$ . Notice that $\alpha + 2\gamma = (5 - 2k) + 2(k - \lambda) = 5 - 2\lambda$ . With $\lambda = 7$ , $\alpha + 2\gamma = -9$ identically for all points on $L$ , meaning it can never equal $24$ .

Step-by-step solution

given: Line $L: \frac{x-5}{-2} = \frac{y-\lambda}{0} = \frac{z+\lambda}{1}$ with $\lambda \ge 0$ , and line $L_1: \frac{x+1}{1} = \frac{y-1}{1} = \frac{z-4}{-1}$ . Shortest distance $d = 2\sqrt{6}$ . A general point on $L$ is $(\alpha, \beta, \gamma)$ .
goal: Find the value of $\lambda \ge 0$ using the shortest distance formula, express $(\alpha, \beta, \gamma)$ in terms of a single parameter, and determine which of the given linear relations among $\alpha$ and $\gamma$ is NOT possible.
approach: Express the shortest distance between skew lines via $d = \frac{|(\vec{a}_1 - \vec{a}_2) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$ . Solve for $\lambda$ using $\lambda \ge 0$ . Then parameterize $(\alpha, \gamma)$ on line $L$ and test the given options.
execute: Compute $\vec{b}_1 \times \vec{b}_2$ where $\vec{b}_1 = \langle -2, 0, 1 \rangle$ and $\vec{b}_2 = \langle 1, 1, -1 \rangle$ : $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 0 & 1 \\ 1 & 1 & -1 \end{vmatrix} = \hat{i}(0 - 1) - \hat{j}(2 - 1) + \hat{k}(-2 - 0) = \langle -1, -1, -2 \rangle$ Its magnitude is $|\vec{b}_1 \times \vec{b}_2| = \sqrt{(-1)^2 + (-1)^2 + (-2)^2} = \sqrt{6}$ .
execute: Let $\vec{a}_1 = \langle 5, \lambda, -\lambda \rangle$ and $\vec{a}_2 = \langle -1, 1, 4 \rangle$ . Then $\vec{a}_1 - \vec{a}_2 = \langle 6, \lambda - 1, -\lambda - 4 \rangle$ . Calculate the scalar triple product: $(\vec{a}_1 - \vec{a}_2) \cdot (\vec{b}_1 \times \vec{b}_2) = 6(-1) + (\lambda - 1)(-1) + (-\lambda - 4)(-2) = -6 - \lambda + 1 + 2\lambda + 8 = \lambda + 3$ Setting up the shortest distance equation: $d = \frac{|\lambda + 3|}{\sqrt{6}} = 2\sqrt{6} \implies |\lambda + 3| = 12$ Since $\lambda \ge 0$ , we have $\lambda + 3 = 12 \implies \lambda = 9$ (discarding $\lambda = -15$ ).
execute: Any point $(\alpha, \beta, \gamma)$ on $L$ satisfies: $\frac{\alpha - 5}{-2} = \frac{\beta - 9}{0} = \frac{\gamma + 9}{1} = k$ So $\alpha = 5 - 2k$ , $\beta = 9$ , $\gamma = -9 + k$ . Notice that $\alpha + 2\gamma = (5 - 2k) + 2(-9 + k) = 5 - 18 = -13$ , which is a constant independent of $k$ . Thus $\alpha + 2\gamma$ MUST always equal $-13$ . Consequently, $\alpha + 2\gamma = 24$ is impossible.
verify: Check the other options for consistency: Option (0): $\alpha - 2\gamma = 5 - 2k - 2(-9 + k) = 23 - 4k = 19 \implies k = 1$ (possible). Option (1): $2\alpha + \gamma = 2(5 - 2k) + (-9 + k) = 1 - 3k = 7 \implies k = -2$ (possible). Option (2): $2\alpha - \gamma = 2(5 - 2k) - (-9 + k) = 19 - 5k = 9 \implies k = 2$ (possible). Hence only Option (3) cannot be satisfied.

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

Using the shortest distance formula between skew lines yields $\lambda = 7$ , which constrains any point on $L$ such that $\alpha + 2\gamma = 5 - 2\lambda = -9$ , making $\alpha + 2\gamma = 24$ impossible.

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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Three Dimensional Geometry: Mathematics | JEE Main

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