In a $∆\mathrm{ABC},$ orthocentre is $\mathrm{H}(2354, 981).$ If $\mathrm{A}(2,1), \mathrm{B}(-10,6)$ then the distance between the orthocentres of $∆\mathrm{HBC}, ∆\mathrm{HAC}$ is

Given vertices of triangle are A=(2, 1), B=(-10, 6) and Orthocentre (H)=(2354, 981)

Clearly Orthocentre of $∆\mathrm{HB}C is A=(2, 1)$

and Orthocentre of $∆\mathrm{HAC} is B=(-10, 6)$

Now $\mathrm{AB}=\sqrt{{(-10-2)}^2+{(6-1)}^2}=\sqrt{144+25}=\sqrt{169}=13$