Let $P$ be a point in the first octant, whose image $Q$ in the plane $x + y = 3$ (that is, the line segment $PQ$ is perpendicular to the plane $x + y = 3$ and the mid-point of $PQ$ lies in the plane $x + y = 3$) lies on the $z$-axis. Let the distance of $P$ from the $x$-axis be $5$. If $R$ is the image of $P$ in the $xy$-plane, then the length of $PR$ is ______.

$\text{Given: } P \text{ in the first octant}$ $Q = \text{image of } P \text{ in the plane } x + y = 3$ $Q \text{ lies on the } z\text{-axis}, \quad \text{distance of } P \text{ from the } x\text{-axis} = 5$ $\text{To find: } |PR| \text{, where } R \text{ is the image of } P$ $\text{in the } xy\text{-plane}$ $1.\ P(x,y,z),\ \text{find the image } Q \text{ in } x+y=3$ $2.\ Q \text{ on the } z\text{-axis} \Rightarrow x, y$ $3.\ d(P, x\text{-axis}) = 5 \Rightarrow z,\ \text{then } R \text{ and } PR$ $\text{Plane } \pi:\ x + y = 3, \qquad \vec{n} = (1, 1, 0)$ $P(x, y, z), \qquad x, y, z > 0$ $\text{Reflection: } PQ \perp \pi, \quad \text{midpoint of } PQ \in \pi$ $F = P - \dfrac{(\vec{n}\cdot P - 3)}{|\vec{n}|^{2}}\,\vec{n}, \qquad |\vec{n}|^{2} = 2$ $F = P - \dfrac{x+y-3}{2}\,(1, 1, 0)$ $Q = 2F - P = P - (x+y-3)(1, 1, 0)$ $Q = (3-y,\ 3-x,\ z)$ $Q \text{ on the } z\text{-axis} \Rightarrow 3-y = 0,\ \ 3-x = 0$ $\Rightarrow x = 3,\quad y = 3$ $\sqrt{y^{2} + z^{2}} = 5$ $9 + z^{2} = 25 \Rightarrow z^{2} = 16$ $z = 4 \quad (\text{since } P \text{ is in the first octant})$ $P(3, 3, 4), \qquad Q(0, 0, 4)$ $M = \dfrac{P+Q}{2} = (1.5, 1.5, 4)$ $x + y = 3 \ \checkmark \Rightarrow Q \text{ is the correct image}$ $R = (3, 3, -4) \quad (\text{image of } P \text{ in the } xy\text{-plane})$ $PR = \sqrt{(3-3)^{2} + (3-3)^{2} + (4-(-4))^{2}} = \sqrt{64} = 8$ $\boxed{\text{Answer: } 8}$