If the straight lines (x-1)/k = (y-2)/2 = (z-3)/3 and (x-2)/3 = (y-3)/k = (z-1)/2 intersect at a poi

Q: Why do we use the scalar triple product? Can't we just equate the parametric coordinates?

You absolutely could! You could set $x_1 = x_2$, $y_1 = y_2$, $z_1 = z_2$ and solve for the parameters $\lambda$ and $\mu$. However, that leads to a system of three equations with two variables, which is often more prone to calculation errors. The scalar triple product is a powerful, elegant shortcut that directly enforces the geometric condition of coplanarity without needing to find the specific intersection point.

Q: Wait, why do we reject the $\frac{5}{2}$ root? Is it mathematically invalid?

It is not mathematically invalid for the lines to intersect at $k = \frac{5}{2}$. However, the question specifically asks for the **integer** value of $k$. In JEE exams, reading the constraints is just as important as solving the equations. Always double-check if the question asks for an integer, a positive value, or a specific range.

Q: What if the lines were parallel? Would the determinant still be zero?

Excellent question! If the lines are parallel, their direction vectors $\vec{d_1}$ and $\vec{d_2}$ are proportional. This makes the second and third rows of the determinant linearly dependent, which forces the determinant to be zero regardless of the points. However, a zero determinant only tells us they are coplanar. If they are parallel, they might be distinct (no intersection) or coincident (infinite intersections). In this specific problem, we can see the direction vectors are not proportional for any $k$, so we don't need to worry about the parallel case.

Question

If the straight lines $\frac{x-1}{k} = \frac{y-2}{2} = \frac{z-3}{3}$ and $\frac{x-2}{3} = \frac{y-3}{k} = \frac{z-1}{2}$ intersect at a point, then the integer $k$ is equal to

Options:

(A) $-5$

(B) $5$

(C) $2$

(D) $-2$

Sigma Percentile · Accepted Answer

### The Dance of Lines in 3D Space: A Geometric Odyssey

Imagine you are standing in a vast, empty room. You have two thin, infinite wires floating in front of you. In the world of 3D geometry, these wires are lines.

Most of the time, if you throw two lines into space, they will be 'skew'—they will never touch, and they will never be parallel. They are like ships passing in the night.

But today, we are looking for a special case. We want these lines to intersect at a single, glorious point. To make this happen, we must impose a strict condition: they must be coplanar.

### Phase 1: Identifying the Players

Every line in 3D space is defined by two things: a point it passes through and a direction in which it points.

Look at our first line, $L_1$:
$$\frac{x-1}{k} = \frac{y-2}{2} = \frac{z-3}{3}$$

By simply glancing at the symmetric form, we can extract its soul. It passes through the point $P_1(1, 2, 3)$ and travels in the direction $\vec{d_1} = (k, 2, 3)$.

Now, look at the second line, $L_2$:
$$\frac{x-2}{3} = \frac{y-3}{k} = \frac{z-1}{2}$$

It passes through $P_2(2, 3, 1)$ and moves with the direction $\vec{d_2} = (3, k, 2)$. These are our actors, floating and waiting for us to find the value of $k$ that forces them to collide.

### Phase 2: The Coplanarity Condition

How do we force two lines to lie on the same plane? If we draw a vector connecting our two known points, $\vec{P_1 P_2}$, this vector must lie in the same plane as the direction vectors $\vec{d_1}$ and $\vec{d_2}$.

If these three vectors—$\vec{P_1 P_2}$, $\vec{d_1}$, and $\vec{d_2}$—are coplanar, the volume of the parallelepiped they form must be zero. This is the magic of the Scalar Triple Product:

$$[\vec{P_1 P_2} \quad \vec{d_1} \quad \vec{d_2}] = 0$$

This is our master key.

### Phase 3: The Determinant Engine

Now, we translate this geometric intuition into the language of algebra. We construct a determinant where the first row is the vector $\vec{P_1 P_2} = (2-1, 3-2, 1-3) = (1, 1, -2)$.

The second and third rows are our direction vectors. The equation becomes:

$$\begin{vmatrix} 1 & 1 & -2 \ k & 2 & 3 \ 3 & k & 2 \end{vmatrix} = 0$$

Expanding this determinant is where many students stumble. We expand along the first row:

$$1 \cdot (2 \cdot 2 - 3 \cdot k) - 1 \cdot (k \cdot 2 - 3 \cdot 3) + (-2) \cdot (k \cdot k - 2 \cdot 3) = 0$$

Let's simplify this carefully:

$$1(4 - 3k) - 1(2k - 9) - 2(k^2 - 6) = 0$$

Expanding further, we get:

$$4 - 3k - 2k + 9 - 2k^2 + 12 = 0$$

### Phase 4: The Final Resolution

We are left with a quadratic equation:

$$-2k^2 - 5k + 25 = 0$$

To make it look friendly, we multiply by $-1$ to get:

$$2k^2 + 5k - 25 = 0$$

We need to split the middle term. We look for two numbers that multiply to $-50$ and add to $5$. Those numbers are $10$ and $-5$.

$$2k^2 + 10k - 5k - 25 = 0$$

Factoring by grouping, we get $2k(k + 5) - 5(k + 5) = 0$, which leads to:

$$(2k - 5)(k + 5) = 0$$

We have two candidates: $k = \frac{5}{2}$ and $k = -5$. The question asks for the **integer** value, so we must reject the fraction.

Thus, **$k = -5$** is our champion. You have successfully navigated the geometry, the algebra, and the constraints. Well done!

Animated Solution for Mathematics - Three Dimensional Geometry: If the straight lines $\frac{x - 1}{k} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and $\frac{x - 2}{3} = \frac{y - 3}{k} = \frac{z - 1}{2}$ intersect at a point, then the integer $k$ is equal to

Select Answer:

Visualized Solution

Identifying Points and Directions

Condition for Coplanarity

Forming the Determinant

Simplifying the Determinant

Expanding along the First Row

Forming the Quadratic Equation

Factorizing to find $k$

Selecting the Integer Value

The Sigma Insight: Shortest Distance Between Two Skew Lines

The Dance of Lines in 3D Space

Phase 1

Phase 2

Phase 3

Phase 4

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