Tetrahedron incenter and circumcenter

As I showed for the previous question , we can calculate the incentre of the tetrahedron to have coordinates $\left(\frac{5\sqrt{233} + 10\sqrt{249} - 5\sqrt{313}}{17 + \sqrt{233}+ \sqrt{249} + \sqrt{313}}\,,\,\frac{12\sqrt{233}+2\sqrt{313}}{17 + \sqrt{233}+ \sqrt{249} + \sqrt{313}}\,,\,\frac{170}{17 + \sqrt{233}+ \sqrt{249} + \sqrt{313}}\right)$ The outcentre of the tetrahedron is the point of intersection of the three planes with equations $\begin{aligned} \mathbf{r} \cdot (\mathbf{a} - \mathbf{b}) & = \; \tfrac12(|\mathbf{a}|^2 -|\mathbf{b}|^2) \\ \mathbf{r} \cdot (\mathbf{a} - \mathbf{c}) & = \; \tfrac12(|\mathbf{a}|^2 -|\mathbf{c}|^2) \\ \mathbf{r} \cdot (\mathbf{a} - \mathbf{d}) & = \; \tfrac12(|\mathbf{a}|^2 -|\mathbf{d}|^2) \end{aligned}$ and hence has coordinates $\big(\tfrac{99}{34},\tfrac{139}{34},\tfrac{99}{34}\big)$ Thus the distance between the outcentre and the incentre is $\boxed{1.083176362}$ .