Animated Solution for Mathematics - Three Dimensional Geometry: Let P be the image of the point $Q(7,-2,5)$ in the line $L:\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ and $R(5,p,q)$ be a point on L. Then the square of the area of $△PQR$ is

Welcome, fellow traveler, to the fascinating world of 3D geometry. Today, we are not just solving a problem; we are exploring the elegant symmetry of space.

Imagine you are standing in a room where a line $L$ acts as a mirror. A point $Q(7,-2,5)$ is floating in the air, and its reflection $P$ appears on the other side of this mirror. We also have a point $R(5,p,q)$ resting on the line $L$.

Our goal is to find the square of the area of the triangle formed by these three points: $P,Q,$ and $R$.

Before we dive into the reflection, let us ground ourselves. We know point $R(5,p,q)$ lies on the line $L$ defined by:

Since $R$ is on the line, its coordinates must satisfy the line's equation. Substituting the $x$-coordinate $5$ into the equation, we get:

Solving these, we find $p+1=6\Rightarrow p=5$ and $q=8$. Thus, our point $R$ is firmly located at $(5,5,8)$.

Now, let us address the reflection. Point $P$ is the image of $Q$ in line $L$, which means the line $L$ is the perpendicular bisector of the segment $PQ$.

Let $M$ be the foot of the perpendicular from $Q$ to $L$. Note that $M$ is the midpoint of $PQ$.

Here is the beauty of the problem: we do not need to find $P$ explicitly. Because $M$ is the midpoint of $PQ$, the area of $△PQR$ is exactly twice the area of $△MQR$. This is because they share the same height from $R$ to the line $PQ$, and the base $PQ$ is twice the base $MQ$.

To find $M$, we express it in terms of a parameter $t$ as $(2t+1,3t-1,4t)$. The vector $\overrightarrow{QM}$ is calculated as:

Since $\overrightarrow{QM}$ is perpendicular to the line's direction vector $\vec{d}=(2,3,4)$, their dot product must be zero:

Expanding this, we get $4t-12+9t+3+16t-20=0$, which simplifies to $29t-29=0$, or $t=1$. Plugging $t=1$ back into the coordinate expression, we find $M=(3,2,4)$.

Now we have $\overrightarrow{MQ}=(7-3,-2-2,5-4)=(4,-4,1)$ and $\overrightarrow{MR}=(5-3,5-2,8-4)=(2,3,4)$. The area of $△MQR$ is $\frac{1}{2}|\overrightarrow{MQ}\times \overrightarrow{MR}|$.

Therefore, the area of $△PQR$ is simply $|\overrightarrow{MQ}\times \overrightarrow{MR}|$. Computing the cross product:

This results in the vector $-19\hat{i}-14\hat{j}+20\hat{k}$. The square of the area is the square of the magnitude of this vector:

We have arrived at the solution. The square of the area of $△PQR$ is 957.