Animated Solution for Mathematics - Three Dimensional Geometry: The lines $x=ay-1=z-2$ and $x=3y-2=bz-2,(ab\ne 0)$ are coplanar, if :

A Journey into 3D Geometry

Imagine you are standing in a vast, empty room. You have two thin, straight wires suspended in the air. The question is simple: can you place a single, flat sheet of paper such that both wires lie entirely on it?

This is the essence of the concept of coplanar lines. In the world of JEE Advanced, this isn't just a visualization exercise; it is a rigorous test of your ability to translate geometric intuition into algebraic precision.

The Trap of Non-Standard Forms

Before we can perform any calculations, we must respect the language of 3D geometry. The equations provided, $x=ay-1=z-2$ and $x=3y-2=bz-2$, are not in their standard symmetric form.

A common mistake is to blindly grab the coefficients of $y$ and $z$ and call them the direction ratios. The standard form is:

where the coefficients of $x,y,$ and $z$ are all $1$.

For $L_1$, we rewrite $x=ay-1=z-2$ as:

Now, the truth is revealed: our point $A$ is $(0,\frac{1}{a},2)$ and our direction vector ${\vec{d}}_1$ is $(1,\frac{1}{a},1)$.

We perform the same surgery on $L_2$, yielding point $B=(0,\frac{2}{3},\frac{2}{b})$ and direction vector ${\vec{d}}_2=(1,\frac{1}{3},\frac{1}{b})$. By standardizing, we have stripped away the confusion and exposed the core geometry.

The Geometry of Coplanarity

Now, how do we force these lines to share a plane? We connect them with a vector $\overrightarrow{AB}=B-A$.

If $\overrightarrow{AB}$, ${\vec{d}}_1$, and ${\vec{d}}_2$ all lie on the same plane, they cannot form a 3D volume. Mathematically, this means their scalar triple product must vanish:

This is the moment where geometry meets algebra. We construct our determinant:

Notice that beautiful zero in the top-left corner? That is a gift from the problem setter. It simplifies our expansion significantly, allowing us to focus on the remaining terms without getting lost in a sea of arithmetic.

The Algebraic Dance

Let us expand along the first row. We have:

This looks intimidating, but stay calm. Look for common factors. We can rewrite $\frac{2}{b}-2$ as $2(\frac{1}{b}-1)$.

Suddenly, the term $(\frac{1}{b}-1)$ appears in both parts of the equation! We factor it out:

Now, simplify the expression inside the square brackets. Distributing the negative sign and the $2$ gives us $-\frac{2}{3}+\frac{1}{a}+\frac{2}{3}-\frac{2}{a}$.

The constants $-\frac{2}{3}$ and $+\frac{2}{3}$ cancel out perfectly, leaving us with $-\frac{1}{a}$. Our equation reduces to the elegant form:

Since the problem guarantees $a eq 0$, we know that $-\frac{1}{a}$ cannot be zero. Therefore, the only way for this product to be zero is if $\frac{1}{b}-1=0$, which implies $b=1$.

And just like that, the complexity dissolves. We have found our condition: $b=1$, with $a$ being any non-zero real number. You have successfully navigated the 3D space and tamed the algebra. Well done!