Animated Solution for Mathematics - Three Dimensional Geometry: The distance of the point $(1,-2,3)$ from the plane $x-y+z=5$ measured parallel to the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$ is:

We are given a point $P(1,-2,3)$ and a plane defined by the equation:

We are tasked with finding the distance from point $P$ to this plane, measured along the direction of the line defined by the vector $\vec{v}=2\hat{i}+3\hat{j}-6\hat{k}$.

To find the intersection point $Q$ where the trajectory meets the plane, we express the line passing through $P$ in parametric form. Using the direction ratios $(2,3,-6)$ and the starting point $(1,-2,3)$, any point $Q$ on this line can be represented as:

Here, $\lambda$ represents the scalar parameter that allows us to traverse the line.

To find the specific point $Q$ that lies on the plane, we substitute these parametric coordinates into the plane equation $x-y+z=5$:

Simplifying the expression:

Solving for the parameter, we find:

Now that we have $\lambda =\frac{1}{7}$, we can determine the distance $PQ$. The distance between point $P$ and point $Q$ is given by the magnitude of the vector $\overrightarrow{PQ}=\lambda \vec{v}$.

The distance $d$ is:

First, calculate the magnitude of the direction vector $\vec{v}=2\hat{i}+3\hat{j}-6\hat{k}$:

Substituting the values of $\lambda$ and $|\vec{v}|$:

The distance from point $P$ to the plane along the given direction is 1.