Animated Solution for Mathematics - Vector Algebra: Let $\alpha ,\beta ,\gamma$ be distinct real numbers. The points with position vectors $\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$, $\beta \hat{i}+\gamma \hat{j}+\alpha \hat{k}$, $\gamma \hat{i}+\alpha \hat{j}+\beta \hat{k}$

Imagine you are standing in a vast $3D$ coordinate system. You are given three points, $A(\alpha ,\beta ,\gamma )$, $B(\beta ,\gamma ,\alpha )$, and $C(\gamma ,\alpha ,\beta )$.

At first glance, they might look like a random collection of variables. However, each point is simply a cyclic shift of the previous one. This is the hallmark of a highly symmetric system, and in geometry, symmetry is almost always a shortcut to elegance.

To understand the nature of the triangle formed by these points, we need to determine the lengths of its sides. We reach for our most trusted tool in $3D$ geometry: the distance formula.

For any two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$, the distance $d$ is given by:

This formula serves as our bridge between algebra and the physical shape of the triangle.

Let us calculate the length of side $AB$. Substituting our coordinates, we get:

Now, let us look at side $BC$. Using the coordinates of $B$ and $C$, we find:

Notice that the terms inside the square root are identical to those in $AB$, just in a different order. Because addition is commutative, $AB$ is exactly equal to $BC$.

Finally, we calculate $CA$:

Again, the terms are identical to the previous calculations.

We have proven that $AB=BC=CA$. A triangle where all three sides are equal is, by definition, an equilateral triangle.

This result is not just a calculation; it is a testament to the power of symmetry. Even when the variables $\alpha ,\beta ,\gamma$ are unknown, the cyclic structure forces the points to maintain a perfect, balanced relationship.

You have just navigated a complex $3D$ problem by simply observing the underlying pattern. Keep this intuition sharp—it is exactly what you need to excel in JEE Advanced.