Animated Solution for Mathematics - Three Dimensional Geometry: If the image of the point $P(1,-2,3)$ in the plane, $2x+3y-4z+22=0$ measured parallel to the line, $\frac{x}{1}=\frac{y}{4}=\frac{z}{5}$ is Q, then PQ is equal to:

Imagine you are standing in a vast, three-dimensional void. You have a point $P(1,-2,3)$ floating in front of you.

Below it, stretching infinitely, is a plane defined by the equation:

In most textbook problems, we are asked to find the image of $P$ by dropping a perpendicular line to the plane. Today, we are looking for a path defined by a specific direction, which highlights the beauty of 3D geometry and the constraints of the space we are working in.

We are given a reference line:

This line acts as our compass. The direction ratios of this line are $⟨1,4,5⟩$, meaning any line parallel to this reference line must share this same direction vector.

Our journey from $P$ to its image $Q$ is a straight line parallel to $⟨1,4,5⟩$. We can express this line parametrically using a parameter $\lambda$:

Any point $M$ on this line can be written as $(1+\lambda ,-2+4\lambda ,3+5\lambda )$. This point $M$ slides along our path until it pierces the plane.

The point $M$ is the intersection of our line and the plane. Because $M$ lies on the plane, it must satisfy the plane's equation.

Substituting our parametric coordinates into the plane equation:

Expanding this expression, we obtain:

Combining the terms, we find:

This is the "magic" value that tells us exactly how far we need to travel along our vector to hit the plane.

With $\lambda =1$, we find the coordinates of $M$:

This is the point on the plane. Since $M$ is the midpoint of the segment $PQ$, the distance $PQ$ is simply twice the distance $PM$.

We calculate the distance $PM$ as follows:

Therefore, the final distance is:

$$PQ=2\sqrt{42}$$

We have successfully navigated the slanted path. Remember, in JEE Advanced, geometry is the key; once you visualize the path, the algebra is simply the tool to reach the solution.