Animated Solution for Mathematics - Three Dimensional Geometry: The distance of the point $(1,1,9)$ from the point of intersection of the line $\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}$ and the plane $x+y+z=17$ is:

Imagine you are standing in a vast, three-dimensional room. You have a straight, infinite path—a line—stretching across the room, and a massive, flat wall—a plane—standing right in its way.

Your goal is to find the exact point where this path pierces the wall and then calculate how far that point is from where you are standing. We are given a point $P(1,1,9)$, a line defined by

and a plane $x+y+z=17$.

A line in 3D space is a collection of infinite points. To capture them all, we introduce a parameter, $\lambda$.

By setting the entire line equation equal to $\lambda$, we express any point $(x,y,z)$ on the line in terms of this single variable:

This is our general point. It acts like a GPS coordinate for every single spot on that line, controlled entirely by the value of $\lambda$.

We look for the intersection point $Q$. Since $Q$ lies on the line, it must have the coordinates $(\lambda +3,2\lambda +4,2\lambda +5)$ for some specific value of $\lambda$.

Because $Q$ also lies on the plane $x+y+z=17$, its coordinates must satisfy the plane's equation. We substitute our parametric expressions into the plane equation:

Simplifying the expression, we combine the $\lambda$ terms to get $5\lambda$ and the constants to get $12$. The equation becomes:

Solving for $\lambda$, we find $5\lambda =5$, which leads us to the elegant result: $\lambda =1$.

With $\lambda =1$ in hand, we find the exact coordinates of our intersection point $Q$. Substituting $\lambda =1$ back into our parametric equations:

Thus, the intersection point is $Q(4,6,7)$.

Finally, we calculate the distance between $P(1,1,9)$ and $Q(4,6,7)$ using the 3D distance formula:

This simplifies to:

The final distance is $\sqrt{38}$.