$\mathrm{I}, {\mathrm{I}}_1,{\mathrm{I}}_2,{\mathrm{I}}_3$ are the incentre and excentres of $∆\mathrm{ABC}$. If $\mathrm{I}(0, 0), {\mathrm{I}}_1(2, 3), {\mathrm{I}}_2(5, 7)$ then the distance between the orthocentres of $∆\mathrm{I} {\mathrm{I}}_1{\mathrm{I}}_3$ and $∆{\mathrm{I}}_1 {\mathrm{I}}_2 {\mathrm{I}}_3$ is

Given incentre (I)=(0, 0) and excentres are ${\mathrm{I}}_1=(2, 3), {\mathrm{I}}_2=(5, 7)$

Since internal angle bisectiors are Perpendicular to external angle bisectors then ‘I’ is orthocenter of ${∆}^{\mathrm{le}} {\mathrm{I}}_1{\mathrm{I}}_2{\mathrm{I}}_3$

$\therefore {\mathrm{I}}_1, {\mathrm{I}}_2, {\mathrm{I}}_3$  Are orthocentres of ${∆}^{\mathrm{le}} \mathrm{I} {\mathrm{I}}_2 {\mathrm{I}}_3, {∆}^{\mathrm{le}} \mathrm{I} {\mathrm{I}}_1 {\mathrm{I}}_3 \mathrm{and} {∆}^{\mathrm{le}} \mathrm{I} {\mathrm{I}}_1 {\mathrm{I}}_2 \mathrm{respectively}.$

$$\begin{gathered} \therefore \mathrm{Orthocentre} \mathrm{of} {∆}^{\mathrm{le}} \mathrm{I} {\mathrm{I}}_1 {\mathrm{I}}_3=(5, 7) \\ \mathrm{Orthocentre} \mathrm{of} {∆}^{\mathrm{le}} \mathrm{I} {\mathrm{I}}_2 {\mathrm{I}}_3=(0, 0) \\ \therefore Required dis\tan ce=\sqrt{25+49}=\sqrt{74} \end{gathered}$$