Animated Solution for Mathematics - Three Dimensional Geometry: Let the image of the point $P(1,2,3)$ in the plane $2x-y+z=9$ be $Q$. If the coordinates of the point $R$ are $(6,10,7)$, then the square of the area of the triangle $PQR$ is _______.

Welcome, fellow traveler of the mathematical landscape. Today, we are not just solving a problem; we are peeling back the layers of a 3D geometric puzzle.

We are given a point $P(1,2,3)$ and a plane defined by $2x-y+z=9$. We are told that $Q$ is the reflection of $P$ across this plane, and we have a third point $R(6,10,7)$.

Our mission is to find the square of the area of the triangle $\Delta PQR$.

Before we rush into complex calculations, let us pause and inspect our surroundings. Is there something special about point $R$?

Let us test it against the plane equation $2x-y+z=9$. Substituting the coordinates of $R(6,10,7)$, we get:

It fits perfectly! This is our first "Aha!" moment. Point $R$ lies exactly on the plane. This means that the distance from $R$ to the plane is zero, and any line segment connecting $R$ to a point on the plane stays within the plane.

Imagine the plane as a mirror. $P$ is an object in front of the mirror, and $Q$ is its reflection. The line segment $PQ$ is perpendicular to the mirror, and the point where it pierces the mirror—let us call it $M$—is the midpoint of $PQ$.

Now, consider the triangle $\Delta PQR$. Because $M$ is the midpoint of $PQ$ and $R$ lies on the plane, we can see that $\Delta PQR$ is composed of two smaller triangles: $\Delta PMR$ and $\Delta QMR$.

Since $PQ$ is the normal to the plane and $RM$ lies within the plane, the angle $\angle PMR$ is exactly $90^{\circ }$. This makes $\Delta PMR$ a right-angled triangle. Because of the symmetry of reflection, $\Delta PMR$ and $\Delta QMR$ are congruent.

Thus, the area of $\Delta PQR$ is simply:

Now, we need the lengths $PM$ and $RM$. $PM$ is the perpendicular distance from $P(1,2,3)$ to the plane $2x-y+z-9=0$.

Using the standard distance formula $d=\frac{|a{x}_1+b{y}_1+c{z}_1+d|}{\sqrt{a^2+b^2+c^2}}$, we calculate:

Next, we find $PR$ using the distance formula between $P(1,2,3)$ and $R(6,10,7)$:

With $PR$ as the hypotenuse and $PM$ as one leg of the right-angled triangle $\Delta PMR$, we find $RM$ using the Pythagorean theorem:

We are almost there. The area of $\Delta PQR$ is $PM\times RM=\sqrt{6}\times \sqrt{99}=\sqrt{594}$.

The question asks for the square of this area. Squaring $\sqrt{594}$ gives us exactly 594.

Look at how the geometry simplified the algebra. We didn't need to find $Q$, we didn't need to deal with complex vectors, and we didn't need to solve a system of equations. By understanding the physical reality of the reflection and the properties of the plane, we turned a daunting 3D problem into a simple exercise in right-angled triangles.