I suppose the notation $S(OAB)$ is the same as $S_{\triangle OAB}$ or $[OAB]$ I haven't gave a thought in it, but I believe it can be solved by using the formula $OH^2=9R^2-(a^2+b^2+c^2)$.

$S_{ABC}=3S_{BOC}(1)$let $AO$ intersect $BC$ at $A'$ from $(1)$ we can get$\frac {AO}{OA'}=2$ so Euler line is parallel to $BC$. in triangle $AOH$ use pythagores theorem and $OH^2=9R^2-(a^2+b^2+c^2)$ formula . with Casi theorem we can calculate the distancse

What does Casi theorem say?

If it is Casey Theorem, then you can find it in the Geometry Formula section.

the statement is wrong, if $\triangle ABC$ is acute, then the distance should be $\frac {a^2}{4\sqrt{9R^2-(a^2+b^2+c^2)}}$. and I believe that the result will be different when $\triangle ABC$ is obtuse.