picking a word for pumping lemma for L = {a^n b^m c^n b^m a^n | m,n≥0}
If you want to apply the pumping lemma for regular languages, you can pick any word in the language whose length is no less than the pumping length. The same applies to the pumping lemma for context-free languages.

However, some of the words in your list may not be in the language.

$a^k c^k$ is in the language if and only if $k=0$.
$a^k b^m c^k $ is in the language if and only if $k=0$ and $m$ is even.
$b^mb^m$ is always in the language.
$b^m c^n b^m$ is in the language if and only if $n=0$.

Yes, a general approach is to check the balance between different symbols. Had the language been $\{a^nc^n\mid n\ge0\}$ or $\{a^n b^m c^n b^m\mid m,n\ge0\}$, you could select $a^kc^k$ towards a contradiction. You cannot if the language is $\{a^n b^m c^n b^m a^n \mid m,n\ge0\}$, since $a^kb^k$ is not a word in the language in the first place. For that language, you can choose $a^kc^ka^k$.

Specifically, here are all the words in $\{a^n b^m c^n b^m a^n \mid m,n\ge0\}$. Note that all of them start and end with the same symbol.

$$\begin{align}
&amp;\epsilon,\\   
&amp;aca,\ a^2c^2a^2,\ a^3c^3a^3,\ a^4c^4a^4,\ \cdots\\
&amp;bb,\ a^1bc^1ba^1,\ a^2bc^2ba^2,\ a^3bc^3ba^3,\ a^4bc^4ba^4, \cdots\\
&amp;b^2b^2,\ a^1b^2c^1b^2a^1,\ a^2b^2c^2b^2a^2,\ a^3b^2c^3b^2a^3,\ a^4b^2c^4b^2a^4,\ \cdots\\
&amp;b^3b^3,\ a^1b^3c^1b^3a^1,\ a^2b^3c^2b^3a^2,\ a^3b^3c^3b^3a^3,\ a^4b^3c^4b^3a^4,\ \cdots\\
&amp;\vdots
\end{align}$$