Animated Solution for Mathematics - Three Dimensional Geometry: The vertices $B$ and $C$ of a triangle $ABC$ lie on the line $\frac{x}{1}=\frac{1-y}{-2}=\frac{z-2}{3}$. The coordinates of $A$ and $B$ are $(1,6,3)$ and $(4,9,\alpha )$ respectively and $C$ is at a distance of 10 units from $B$. The area (in sq. units) of $△ABC$ is :

Welcome, future engineer. Today, we are not just solving a problem; we are painting a picture in three-dimensional space. Imagine you are standing in a room with a line running across the floor and a point $A$ floating in the air above it.

You have a triangle $ABC$ where $B$ and $C$ are anchored to that line, and $A$ is your apex. The problem asks for the area of this triangle. It sounds simple, but the beauty lies in the precision of our approach.

Before we do any heavy lifting, we must ensure our tools are sharp. The problem gives us the line $L$ as:

Many students rush here, grabbing the denominator $⟨1,-2,3⟩$ as the direction vector. Stop! Look at the $y$-term: $1-y$.

The standard form requires the variable to be positive, in the form $y-y_1$. By multiplying the numerator and denominator of that middle term by $-1$, we transform it into $\frac{y-1}{2}$. Now, the direction vector $\vec{v}$ reveals its true self: $\vec{v}=\hat{i}+2\hat{j}+3\hat{k}$.

We are given point $B(4,9,\alpha )$ on the line. You might feel the urge to solve for $\alpha$ immediately. While we can easily find that $\alpha =14$ by substituting $B$ into the line equation, ask yourself: do we actually need it?

The area of a triangle is $\frac{1}{2}\times \text{base}\times \text{height}$. We are given the base $BC=10$. The height is the perpendicular distance from $A$ to the line.

Notice that the coordinates of $B$ do not change the height of the triangle relative to the line. $\alpha$ is a distractor—a test of your focus. We proceed to the real challenge: finding the height.

To find the height, we must drop a perpendicular from point $A(1,6,3)$ to the line $L$. Let the foot of this perpendicular be $D$. Since $D$ lies on the line, we can define its coordinates using a parameter $\lambda$.

By equating the line equation to $\lambda$, we get the general coordinates of any point on the line: $D=(\lambda ,2\lambda +1,3\lambda +2)$.

Now, we construct the vector $\overrightarrow{AD}$, which connects our apex $A$ to the line. Subtracting the position vectors, we get:

This simplifies to:

Here is where the physics and math collide in perfect harmony. Because $AD$ is the perpendicular height, the vector $\overrightarrow{AD}$ must be orthogonal to the line's direction vector $\vec{v}=\hat{i}+2\hat{j}+3\hat{k}$. The condition for orthogonality is that their dot product must be zero: $\overrightarrow{AD}\cdot \vec{v}=0$.

Let us perform this calculation with care:

Expanding this, we get:

Combining the terms:

This yields $\lambda =1$. Substituting $\lambda =1$ back into our expression for $D$, we find the foot of the perpendicular: $D=(1,2(1)+1,3(1)+2)=(1,3,5)$.

We are at the finish line. The height $h$ is the length of segment $AD$. Using the distance formula between $A(1,6,3)$ and $D(1,3,5)$:

Finally, the area of $△ABC$ is:

Look at what you have achieved. You navigated the trap of the line equation, ignored the distractor, utilized the power of the dot product, and arrived at the solution with precision. This is the essence of JEE Advanced—not just calculating, but visualizing and understanding the geometry of the world around you.