1. Set coordinates of the point and its image

Let $P=(x,y,z)$ with $x,y,z>0$ since $P$ is in the first octant.

Its image $Q$ in the plane $x+y=3$ is obtained by reflection across that plane.

We are given that $Q$ lies on the $Z$-axis, so $Q=(0,0,q)$ for some $q$.

2. Use reflection formula across the plane

The plane is $x+y-3=0$ with normal vector $\vec n=(1,1,0).$

Reflection of point $P=(x,y,z)$ in the plane $x+y-3=0$ is $Q=P-2\frac{x+y-3}{1^2+1^2}(1,1,0).$

Since $1^2+1^2=2$, this becomes $Q=(x,y,z)-(x+y-3)(1,1,0).$

So, $Q=(x-(x+y-3),\ y-(x+y-3),\ z).$

Hence, $Q=(3-y,\ 3-x,\ z).$

But $Q$ lies on the $Z$-axis, so its $x$- and $y$-coordinates are zero: $3-y=0, \quad 3-x=0.$ Thus, $x=3,\quad y=3.$

Therefore, $P=(3,3,z).$

3. Use distance of $P$ from the $X$-axis

Distance of a point $(x,y,z)$ from the $X$-axis is $\sqrt{y^2+z^2}.$

Given this distance is $5$, $\sqrt{3^2+z^2}=5.$ So, $9+z^2=25$ $z^2=16$ Since $P$ is in the first octant, $z>0$, hence $z=4.$

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

4. Find point $R$, the image of $P$ in the $XY$-plane

Reflection in the $XY$-plane changes $z$ to $-z$, so $R=(3,3,-4).$

5. Compute $PR$

Distance between $P=(3,3,4)$ and $R=(3,3,-4)$ is $PR=\sqrt{(3-3)^2+(3-3)^2+(4-(-4))^2}$ $=\sqrt{0+0+8^2}=8.$

6. Final answer

$\boxed{8}$