Vertices of a triangle are $\mathrm{A}(0, 0), \mathrm{B}(6, 8), \mathrm{C}(8,-6).$ The distance between orthocentre $\mathrm{̒} \mathrm{H} \mathrm{̓}$ and circumcentre $\text{'}\mathrm{S}\text{'}$ is

Given vertices of triangle are

A=(0, 0), B=(6, 8), C=(8, -6)

$$\begin{gathered} \mathrm{Now} AB=\sqrt{{\left(6\right)}^2+{\left(8\right)}^2}=\sqrt{100} \\ \mathrm{BC}=\sqrt{{(6-8)}^2+{(8+6)}^2}=\sqrt{4+196}=\sqrt{200} \\ \mathrm{CA}=\sqrt{{\left(8\right)}^2+{(-6)}^2}=\sqrt{100} \\ \mathrm{Now} {\mathrm{BC}}^2=A{B}^2+{\mathrm{CA}}^2 \end{gathered}$$

$\therefore$ Triangle right angled triangle at vertex A

$\therefore$Orthocentre (H)=(0, 0)

Circumcentre (S)=mid point of $\overline{BC}$

$$\begin{gathered} \Rightarrow \mathrm{S}=\left(\frac{14}{2}, \frac{2}{2}\right)=(7, 1) \\ \mathrm{Now} \mathrm{H}.\mathrm{S}=\sqrt{{\left(7\right)}^2+{\left(1\right)}^2}=\sqrt{50} \end{gathered}$$