Animated Solution for Mathematics - Three Dimensional Geometry: The distance of the point $(-1,9,-16)$ from the plane $2x+3y-z=5$ measured parallel to the line $\frac{x+4}{3}=\frac{2-y}{4}=\frac{z-3}{12}$ is

Beyond the Perpendicular

Welcome, student. Today, we are not just solving a problem; we are embarking on a journey through 3D space.

Often, when we see a point and a plane, our brain automatically screams, 'Perpendicular distance formula!' But in the JEE Advanced arena, the examiners love to test your conceptual clarity by introducing constraints.

Today, we are tasked with finding the distance of the point $P(-1,9,-16)$ from the plane $2x+3y-z=5$, but with a twist: the distance is measured parallel to the line $L_1:\frac{x+4}{3}=\frac{2-y}{4}=\frac{z-3}{12}$. This is not a shortcut problem; it is a path-finding problem.

The Standardization Trap

Before we move, we must respect the geometry. Look at the line $L_1$. The middle term is $\frac{2-y}{4}$.

If you extract the direction ratios as $(3,4,12)$, you have already fallen into the trap. The standard form requires the coefficient of the variable to be positive $1$.

We must rewrite this as $\frac{y-2}{-4}$. Now, the direction ratios are clearly $(3,-4,12)$. This vector, let us call it $\vec{v}=3\hat{i}-4\hat{j}+12\hat{k}$, is the compass that guides our path. It dictates the direction in which we must travel from point $P$ to reach the plane.

The Parametric Journey

Imagine standing at point $P(-1,9,-16)$. You are holding a laser pointer, and you align it with the direction vector $\vec{v}$.

You fire the laser until it hits the plane at a point $Q$. This line, $L_2$, is our path. Since it passes through $P$ and follows the direction $\vec{v}$, we can write its equation in parametric form:

Here, $\lambda$ is our parameter. It is the variable that allows us to 'walk' along the line. Any point $Q$ on this line can be represented as $(3\lambda -1,-4\lambda +9,12\lambda -16)$. This is the beauty of parametric geometry—we have reduced a 3D point to a single variable $\lambda$.

The Intersection

Now, we seek the point $Q$ where our laser hits the plane. Since $Q$ lies on the plane $2x+3y-z=5$, its coordinates must satisfy the plane's equation.

This is the moment of truth. We substitute our parametric expressions into the plane equation:

Take a deep breath. Let us expand this carefully. We get $6\lambda -2-12\lambda +27-12\lambda +16=5$.

Combining the $\lambda$ terms, we have $(6-12-12)\lambda =-18\lambda$. Combining the constants, we have $-2+27+16=41$.

So, the equation simplifies to $-18\lambda +41=5$. This leads us to $-18\lambda =-36$, which gives us the elegant result: $\lambda =2$.

The Final Victory

We have found our parameter! Substituting $\lambda =2$ back into our expression for $Q$, we find the coordinates of the intersection point:

So, $Q$ is $(5,1,8)$. The final step is simply the distance between $P(-1,9,-16)$ and $Q(5,1,8)$. Using the distance formula:

And there it is. The distance is 26. You navigated the trap, standardized the line, solved the intersection, and arrived at the solution. This is the power of systematic thinking. Keep this clarity, and no problem in JEE will ever be too complex for you.