Animated Solution for Mathematics - Three Dimensional Geometry: The plane passing through the point (4, -1, 2) and parallel to the lines $\frac{x+2}{3}=\frac{y-2}{-1}=\frac{z+1}{2}$ and $\frac{x-2}{1}=\frac{y-3}{2}=\frac{z-4}{3}$ also passes through the point :

Imagine you are standing in a vast, empty 3D coordinate space. You are holding a perfectly flat, infinite sheet of glass. This sheet is our plane.

We are told this glass is parallel to two specific lines, and it must pass through a single, fixed point: $P(4,-1,2)$. To define this plane, we need to know its orientation—its 'tilt'—and one point it anchors to.

In the world of 3D geometry, the orientation of a plane is defined by its normal vector, $\vec{n}$, a vector that stands perfectly perpendicular to the surface, like a flagpole stuck into the ground.

We are given two lines with direction vectors $\overrightarrow{d_1}=3\hat{i}-\hat{j}+2\hat{k}$ and $\overrightarrow{d_2}=\hat{i}+2\hat{j}+3\hat{k}$. Because our plane is parallel to these lines, these direction vectors must lie parallel to the surface of our glass sheet.

If you have two vectors that are parallel to a plane, their cross product, $\vec{n}=\overrightarrow{d_1}\times \overrightarrow{d_2}$, will naturally point in the direction of the normal. This is the magic of the cross product: it creates a vector that is simultaneously perpendicular to both inputs.

To find this, we set up the determinant:

Expanding this, we get $\vec{n}=\hat{i}(-3-4)-\hat{j}(9-2)+\hat{k}(6-(-1))$, which simplifies to $\vec{n}=-7\hat{i}-7\hat{j}+7\hat{k}$.

Now, look at that vector: $(-7,-7,7)$. In geometry, the magnitude of the normal vector does not change the orientation of the plane.

We can divide by $-7$ to make our calculations easier, giving us the direction ratios $(1,1,-1)$. This is our simplified normal vector. It is much cleaner to work with, and it represents the exact same tilt as the original.

With our normal vector $\vec{n}=(1,1,-1)$ and our anchor point $P(4,-1,2)$, we use the point-normal form of the plane equation: $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$.

Substituting our values, we get:

Expanding this, we have $x-4+y+1-z+2=0$, which simplifies beautifully to:

This is the mathematical soul of our plane. Every point $(x,y,z)$ that satisfies this equation lies on our sheet of glass.

Finally, we test our result. We check the point $(1,1,1)$ by plugging it into our equation: $1+1-1-1=0$.

Since the result is zero, we have confirmed that the point $(1,1,1)$ lies perfectly on our plane. You have successfully navigated the 3D landscape, used the cross product to define the orientation, and verified your result.

This is the elegance of JEE mathematics—taking a complex spatial problem and reducing it to a simple, satisfying balance. The final equation of the plane is $x+y-z-1=0$.