Working of PDA for $\{a^m b^n c^k \mid m=n \text{ or } n=k\}$
The PDA first guesses whether $m=n$ or $n=k$. According to the guess, it either just checks that $m=n$, or just checks that $n=k$.
What your heuristic argument suggests is that this language cannot be accepted by a deterministic PDA. You can likely show this by adapting the proof here.