Distance between Image Points of a Spatial Point in Octant One

Let $P = (x_1, y_1, z_1)$ be a point in the first octant, which implies that $x_1 > 0$, $y_1 > 0$, and $z_1 > 0$.

Since the image $Q$ of $P$ lies on the $z$-axis, its coordinates can be written as: $Q = (0, 0, z_0)$

The line segment $PQ$ is perpendicular to the plane $x + y = 3$, whose normal vector is $\vec{n} = (1, 1, 0)$. The vector $\vec{PQ}$ is given by: $\vec{PQ} = Q - P = (-x_1, -y_1, z_0 - z_1)$

Since $\vec{PQ}$ is parallel to the normal vector $\vec{n} = (1, 1, 0)$, we must have: $z_0 - z_1 = 0 \implies z_0 = z_1$ $\frac{-x_1}{1} = \frac{-y_1}{1} \implies x_1 = y_1$

The midpoint $M$ of $PQ$ is: $M = \left( \frac{x_1 + 0}{2}, \frac{y_1 + 0}{2}, \frac{z_1 + z_0}{2} \right) = \left( \frac{x_1}{2}, \frac{x_1}{2}, z_1 \right)$

Given that $M$ lies on the plane $x + y = 3$: $\frac{x_1}{2} + \frac{x_1}{2} = 3 \implies x_1 = 3$

Thus, $y_1 = 3$, and the coordinates of $P$ are $(3, 3, z_1)$.

Next, the perpendicular distance of the point $P(3, 3, z_1)$ from the $x$-axis is given by $\sqrt{y_1^2 + z_1^2}$. We are given that this distance is equal to $5$: $\sqrt{3^2 + z_1^2} = 5$ $9 + z_1^2 = 25 \implies z_1^2 = 16$

Since $P$ is in the first octant ($z_1 > 0$), we have: $z_1 = 4$

Hence, $P = (3, 3, 4)$.

Since $R$ is the reflection image of $P(3, 3, 4)$ in the $xy$-plane ($z = 0$), its coordinates are: $R = (3, 3, -4)$

The total length of the segment $PR$ is: $PR = \sqrt{(3 - 3)^2 + (3 - 3)^2 + (4 - (-4))^2} = \sqrt{0 + 0 + 8^2} = 8$