smc

We connect all the medians of the triangle. We know that the ratio of the vertice to orthocenter / median is in a $2:3$ ratio. For convenience, call the orthocenter $O$, and label the triangle $△ABC$ such that the median from $A$ to $BC$ is 3 and the median from $B$ to $AC$ is 6. Then, $BO$ is 4 and $AO$ is 2. Let us call the portion of the third median that goes from $O$ to $AB$ have length $x$, and and $OC$ have length $2x$. Note that the median from $O$ to $AB$ in $△AOB$ is equal to $x$. Note that the medians split the triangle into 6 triangles of equal area, so $△AOB$ has area equal to $\frac{1}{3}$ of $△ABC=\sqrt{15}$. Let $AB=2s$. Using Herons, we get: $$15=(3+s)(s-1)(s+1)(3-2)$$ $$=(9-s^2)(s^2-1)$$ We can see that $s^2=4$, meaning that $s$ is 2 and $AB=4$. We can then use Steward's to find the length of the median from $O$, since we know the median cuts $AB$ into segments each of length $2$. We get: $$(2^2)(4)+4x^2=(4^2)(2)+(2^2)(2)$$ $$16+4x^2=32+8$$ $$4x^2=24$$ $$x^2=6$$ $$x=\sqrt{6}$$ Since the length of the actual median from $C$ is equal to $3$, we have that the answer is $\boxed{3\sqrt{6}}$. * If anyone has a better method of either finding $AB$ or the median of $AOB$ from $O$, please feel free to edit