The distance between the orthocenter and circumcentre of the triangle formed by the points $\left(1,2,3\right),\left(3,-1,5\right),\left(4,0,-3\right)$

Vertices of the given triangle are $\left(1,2,3\right),\left(3,-1,5\right),\left(4,0,-3\right)$

Let us find the sides of the triangle

$$\begin{gathered} AB=\sqrt{4+9+4}=\sqrt{17} \\ BC=\sqrt{1+1+64}=\sqrt{66} \\ CA=\sqrt{9+4+36}=\sqrt{49} \end{gathered}$$

observing the above sides, ${\left(BC\right)}^2={\left(AB\right)}^2+{\left(AC\right)}^2$

Hence, the triangle is Right angled triangle

The vertex at right angle is ortho center, hace the orthocenter of the above triangle is $A\left(1,2,3\right)$

The distance between orhto center and circum center is equal to the half of the length of the median

Hence, the required distance is $\frac{\sqrt{66}}{2}=\sqrt{\frac{33}{2}}$ units