5. In the Cartesian coordinate system, it is known that $A(13,43)$, $B(-36,19 \sqrt{2})$, $C(18,-11 \sqrt{14})$. Then the coordinates of the orthocenter $H$ of $\triangle A B C$ are $\qquad$.
5. $(-5,43+19 \sqrt{2}-11 \sqrt{14})$.

Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(x_{3}, y_{3}\right)$.
Since $x_{1}^{2}+y_{1}^{2}=x_{2}^{2}+y_{2}^{2}=x_{3}^{2}+y_{3}^{2}=2018$, the circumcenter of $\triangle A B C$ is the origin $O$.
Therefore, $H\left(x_{1}+x_{2}+x_{3}, y_{1}+y_{2}+y_{3}\right)$.
Thus, $H(-5,43+19 \sqrt{2}-11 \sqrt{14})$.