Decide whether the Language is regular {a^i b^j c^k|i ≥ 0, j ≥ 0, k ≥ 0}
You're using the lemma completely wrong. It states that

If language $L$ is regular, then there is $p$ such that for every string $s$ in $L$ where $|s|&gt;p$ there is a decomposition $xyz$ with $|y|&gt;0$ and $|xy|\le p$ for which $xy^i z \in L$ for all $i \ge 0$.

When you want to use it to prove language not to be regualr, you use contrapositive implication (making $\neg B \implies \neg A$ from $A \implies B$ ).

You get if for every $p$ there is a string $s$ with $|s| &gt; p$ where for every decomposition $xyz$ with $|y|&gt;0$ and $|xy| \le p$ there is a $i \ge 0$ such that $xy^i z \notin L$ then $L$ is not regular.

Now for your actual question, what you seemed to be doing is trying to prove the first implication, which would tell you nothing. It's important to note that the lemma is implication not equivalence, thus proving that a language has the pumping property does not mean it is regular. Conversely you cannot prove such languages to not be regular by pumping lemma.

If you wanted to prove that the language is not regular, you would need to make a string $s$ longer than $p$ for every $p$ from the language $L$ (for example by taking string $a^p b$ or $a^{p+1}$ or similar strings) and then you would have to make sure that there is an $i$ for all of the decompositions $xyz$ such that $xy^i z \notin L$. You can find many examples both on this site and on the Internet, but it is impossible here because the language is in fact regular.

To take your example more closely, if you wanted to prove the language to not be regular with the pumping lemma, your first mistake was chosing $p$, because you need to prove the statement for every $p$.