Animated Solution for Mathematics - Three Dimensional Geometry: If the straight lines $x=1+s,y=-3-\lambda s,z=1+\lambda s$ and $x=\frac{t}{2},y=1+t,z=2-t$, with parameters $s$ and $t$ respectively, are co-planar, then $\lambda$ equals.

Understanding Co-planarity

Welcome, future engineer. Today, we are not just solving a problem; we are exploring the architecture of 3D space.

When we talk about two lines being 'co-planar', we are asking a fundamental question: Can these two distinct paths exist on the same flat sheet of paper, or are they destined to pass by each other in the infinite void of three-dimensional space? This is the essence of the JEE Advanced approach—moving beyond rote calculation to visualize the geometric soul of the problem.

Decoding the Lines

Let us begin by deconstructing our lines. We are given two lines in parametric form.

When you see equations like $x=1+s$, $y=-3-\lambda s$, and $z=1+\lambda s$, do not panic. These are simply instructions on how to travel along the line. The constants represent a fixed point on the line, and the coefficients of the parameter $s$ represent the direction in which the line is heading.

For our first line, we extract the point $A(1,-3,1)$ and the direction vector $\overrightarrow{d_1}=(1,-\lambda ,\lambda )$. Similarly, for the second line, we find point $B(0,1,2)$ and the direction vector $\overrightarrow{d_2}=(\frac{1}{2},1,-1)$.

We have now successfully decoded the DNA of these two lines. We know where they start, and we know where they are pointing.

The Bridge Between Worlds

Now, we need to connect these two lines. Imagine standing at point $A$ on the first line and looking toward point $B$ on the second line. The vector connecting them, $\overrightarrow{AB}$, is our bridge.

We calculate this by subtracting the coordinates of $B$ from $A$:

This vector $\overrightarrow{AB}$ is crucial. If the two lines are to lie on the same plane, then the bridge connecting them ($\overrightarrow{AB}$) and the two direction vectors ($\overrightarrow{d_1}$ and $\overrightarrow{d_2}$) must all reside on that same flat surface. If they are all on the same plane, they are linearly dependent.

The Scalar Triple Product

Here is where the magic happens. In vector algebra, the volume of a parallelepiped formed by three vectors $\vec{a},\vec{b},\vec{c}$ is given by the scalar triple product $[\vec{a},\vec{b},\vec{c}]$.

If these three vectors are co-planar, they cannot form a 3D shape with volume. The volume must collapse to zero. Therefore, the condition for co-planarity is:

We translate this into a determinant. This is the moment where the abstract geometry becomes a concrete algebraic task. We set up our determinant using the components we found earlier:

The Final Calculation

Do not let the determinant intimidate you. We expand along the first row. It is a systematic process of breaking down the larger problem into smaller, manageable pieces:

As we simplify, watch how the terms begin to resolve. The first minor gives us $(\lambda -\lambda )=0$. The second term becomes $4(-1-\frac{\lambda }{2})$, and the third term becomes $-1(1+\frac{\lambda }{2})$.

Combining these, we get:

Grouping the constants and the $\lambda$ terms, we arrive at $-5-\frac{5\lambda }{2}=0$. Solving for $\lambda$ is now straightforward:

We have arrived at our destination: $\lambda =-2$. By understanding the geometric requirement of co-planarity and translating it into the language of scalar triple products, we have tamed the complexity of 3D space.

Remember, in JEE Advanced, the math is just the tool; the true skill is the ability to visualize the geometry behind the equations. Keep practicing, keep visualizing, and keep pushing forward.