JEE MainMathematics

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Three Dimensional Geometry: Mathematics | JEE Main

If a plane passes through the points

(-1, \mathrm{k}, 0), (2, \mathrm{k}, -1), (1, 1, 2)

and is parallel to the line

\frac{\mathrm{x}-1}{1} = \frac{2\mathrm{y}+1}{2} = \frac{\mathrm{z}+1}{-1}

, then the value of

\frac{\mathrm{k}^2+1}{(\mathrm{k}-1)(\mathrm{k}-2)}

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.

\frac{17}{5}

\frac{13}{6}

\frac{6}{13}

\frac{5}{17}

Step-by-step solutionView

Correct answer

The parameter $k = -2$ , giving an expression value of $\frac{13}{6}$ .

Option analysis

Why each option works or fails

A ·

\frac{17}{5}

Solving the coplanarity determinant with a sign error leading to $k = 4$ or finding $k$ incorrectly from vector directions. Ensure direction ratios of the line are correctly written in standard form $\frac{y - y_0}{b}$ (giving direction ratio 1 for $y$ , not 2) and carefully expand the $3 \times 3$ determinant.

B ·

\frac{13}{6}

None. The student correctly extracted the line direction vector, established coplanarity with the two displacement vectors in the plane, and evaluated $\frac{k^2+1}{(k-1)(k-2)}$ for $k = -2$ . None.

C ·

\frac{6}{13}

Inverting the final expression to $\frac{(k-1)(k-2)}{k^2+1}$ instead of $\frac{k^2+1}{(k-1)(k-2)}$ . Check that the required ratio has the numerator and denominator in the correct positions before substituting $k$ .

D ·

\frac{5}{17}

Combining an erroneous value of $k$ with an inverted fraction evaluation. Maintain the correct orientation of the ratio requested in the question stem and re-verify the determinant expansion.

Reviewed route

Solution

StepWorking

01given

Plane passes through points $A(-1, k, 0)$ , $B(2, k, -1)$ , and $C(1, 1, 2)$ . It is parallel to the line $\frac{x-1}{1} = \frac{2y+1}{2} = \frac{z+1}{-1}$ .

02goal

Find the value of $k$ and then calculate $\frac{k^2+1}{(k-1)(k-2)}$ .

03approach

First, find the direction ratios of the line by rewriting $\frac{2y+1}{2}$ as $\frac{y+1/2}{1}$ , giving direction vector $\vec{b} = \langle 1, 1, -1 \rangle$ . Two vectors in the plane are $\vec{AB} = \langle 3, 0, -1 \rangle$ and $\vec{AC} = \langle 2, 1-k, 2 \rangle$ . Since the line is parallel to the plane, the normal vector to the plane $\vec{n} = \vec{AB} \times \vec{AC}$ must be perpendicular to $\vec{b}$ , meaning $\vec{b} \cdot (\vec{AB} \times \vec{AC}) = 0$ , which corresponds to the scalar triple product $[\vec{b}, \vec{AB}, \vec{AC}] = 0$ .

04execute

The line vector is $\vec{b} = \langle 1, 1, -1 \rangle$ . Vector $\vec{u} = \vec{AB} = \langle 2 - (-1), k - k, -1 - 0 \rangle = \langle 3, 0, -1 \rangle$ . Vector $\vec{v} = \vec{CB} = \langle 2 - 1, k - 1, -1 - 2 \rangle = \langle 1, k-1, -3 \rangle$ . Since the line is parallel to the plane containing $\vec{u}$ and $\vec{v}$ , these three vectors must be coplanar: $\begin{vmatrix} 1 & 1 & -1 \\ 3 & 0 & -1 \\ 1 & k-1 & -3 \end{vmatrix} = 0$ . Expanding along the first row: $1(0 - (-1)(k-1)) - 1(3(-3) - (-1)(1)) + (-1)(3(k-1) - 0) = 0 \implies (k-1) - (-9+1) - 3(k-1) = 0 \implies -2(k-1) + 8 = 0 \implies k - 1 = 4 \implies k = 5$ .

05execute

Substitute $k = 5$ into the required expression: $\frac{k^2+1}{(k-1)(k-2)} = \frac{5^2 + 1}{(5-1)(5-2)} = \frac{26}{4 \times 3} = \frac{26}{12} = \frac{13}{6}$ .

✓verify

Verify the normal vector with $k=5$ : $\vec{u} = \langle 3, 0, -1 \rangle$ , $\vec{v} = \langle 1, 4, -3 \rangle$ . Normal vector $\vec{n} = \vec{u} \times \vec{v} = \langle 4, 8, 12 \rangle \propto \langle 1, 2, 3 \rangle$ . Check orthogonality with $\vec{b} = \langle 1, 1, -1 \rangle$ : $\vec{n} \cdot \vec{b} = 1(1) + 2(1) + 3(-1) = 0$ . Everything holds consistently.

Hints that build this answer step by step

What is the direction vector $\vec{d}$ of the line $\frac{x-1}{1} = \frac{2y+1}{2} = \frac{z+1}{-1}$ in standard form?
$\langle 1, 1, -1 \rangle$
Let the points on the plane be $A(-1, k, 0)$ , $B(2, k, -1)$ , and $C(1, 1, 2)$ . Since the plane is parallel to the line, how are $\vec{AB}$ , $\vec{AC}$ , and $\vec{d}$ related?
The scalar triple product $[\vec{AB}, \vec{AC}, \vec{d}] = 0$ , because all three vectors are parallel to the plane.
Evaluate the determinant equation $\begin{vmatrix} 3 & 0 & -1 \\ 2 & 1-k & 2 \\ 1 & 1 & -1 \end{vmatrix} = 0$ to find $k$ .
$k = -2$
What is the value of $\frac{k^2+1}{(k-1)(k-2)}$ when $k = 5$ ?
$\frac{13}{6}$

✓ 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 direction ratio along y equal to 1 and not 2?

The standard symmetric form of a line equation requires the coefficient of y to be +1: $\frac{2y+1}{2} = \frac{2(y+1/2)}{2} = \frac{y+1/2}{1}$ , so the denominator in standard form is 1.

Answer

The parameter $k = -2$ , giving an expression value of $\frac{13}{6}$ .

Why each option works or fails

A: $\frac{17}{5}$ - Solving the coplanarity determinant with a sign error leading to $k = 4$ or finding $k$ incorrectly from vector directions. Ensure direction ratios of the line are correctly written in standard form $\frac{y - y_0}{b}$ (giving direction ratio 1 for $y$ , not 2) and carefully expand the $3 \times 3$ determinant.
B · correct: $\frac{13}{6}$ - None. The student correctly extracted the line direction vector, established coplanarity with the two displacement vectors in the plane, and evaluated $\frac{k^2+1}{(k-1)(k-2)}$ for $k = -2$ . None.
C: $\frac{6}{13}$ - Inverting the final expression to $\frac{(k-1)(k-2)}{k^2+1}$ instead of $\frac{k^2+1}{(k-1)(k-2)}$ . Check that the required ratio has the numerator and denominator in the correct positions before substituting $k$ .
D: $\frac{5}{17}$ - Combining an erroneous value of $k$ with an inverted fraction evaluation. Maintain the correct orientation of the ratio requested in the question stem and re-verify the determinant expansion.

Step-by-step solution

given: Plane passes through points $A(-1, k, 0)$ , $B(2, k, -1)$ , and $\frac{13}{6}$ . It is parallel to the line $\frac{x-1}{1} = \frac{2y+1}{2} = \frac{z+1}{-1}$ .
goal: Find the value of $k$ and then calculate $\frac{k^2+1}{(k-1)(k-2)}$ .
approach: First, find the direction ratios of the line by rewriting $\frac{2y+1}{2}$ as $\frac{y+1/2}{1}$ , giving direction vector $\vec{b} = \langle 1, 1, -1 \rangle$ . Two vectors in the plane are $\vec{AB} = \langle 3, 0, -1 \rangle$ and $\vec{AC} = \langle 2, 1-k, 2 \rangle$ . Since the line is parallel to the plane, the normal vector to the plane $\vec{n} = \vec{AB} \times \vec{AC}$ must be perpendicular to $\vec{b}$ , meaning $\vec{b} \cdot (\vec{AB} \times \vec{AC}) = 0$ , which corresponds to the scalar triple product $[\vec{b}, \vec{AB}, \vec{AC}] = 0$ .
execute: The line vector is $\vec{b} = \langle 1, 1, -1 \rangle$ . Vector $\vec{u} = \vec{AB} = \langle 2 - (-1), k - k, -1 - 0 \rangle = \langle 3, 0, -1 \rangle$ . Vector $\vec{v} = \vec{CB} = \langle 2 - 1, k - 1, -1 - 2 \rangle = \langle 1, k-1, -3 \rangle$ . Since the line is parallel to the plane containing $\vec{u}$ and $\vec{v}$ , these three vectors must be coplanar: $\begin{vmatrix} 1 & 1 & -1 \\ 3 & 0 & -1 \\ 1 & k-1 & -3 \end{vmatrix} = 0$ . Expanding along the first row: $1(0 - (-1)(k-1)) - 1(3(-3) - (-1)(1)) + (-1)(3(k-1) - 0) = 0 \implies (k-1) - (-9+1) - 3(k-1) = 0 \implies -2(k-1) + 8 = 0 \implies k - 1 = 4 \implies k = 5$ .
execute: Substitute $k = 5$ into the required expression: $\frac{k^2+1}{(k-1)(k-2)} = \frac{5^2 + 1}{(5-1)(5-2)} = \frac{26}{4 \times 3} = \frac{26}{12} = \frac{13}{6}$ .
verify: Verify the normal vector with $k=5$ : $\vec{u} = \langle 3, 0, -1 \rangle$ , $\vec{v} = \langle 1, 4, -3 \rangle$ . Normal vector $\vec{n} = \vec{u} \times \vec{v} = \langle 4, 8, 12 \rangle \propto \langle 1, 2, 3 \rangle$ . Check orthogonality with $\vec{b} = \langle 1, 1, -1 \rangle$ : $\vec{n} \cdot \vec{b} = 1(1) + 2(1) + 3(-1) = 0$ . Everything holds consistently.

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

The parameter $k = -2$ , giving an expression value of $\frac{13}{6}$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

✓verify

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