First, Understand Pumping Lemma. Clear, Correct & Complete Understating is Needed.
Here is the $\color{red}{\text{Complete “Pumping Lemma for Regular Languages” Playlist}}$, with Proof, Questions, Variations & Results etc..
Complete Pumping Lemma Playlist, with A Lot of Questions: https://www.youtube.com/playlist?list=PLIPZ2_p3RNHjGbysj9OvLTfL2qhsTdsbr
After watching the above playlist, this question & all questions will become Extremely Simple. So, Let’s begin!!
Pumping lemma states a Deep Property that All Regular languages share. By Showing that a language does not have the property stated by Pumping lemma, we are guaranteed that It is Not Regular.
Pumping Lemma(PL) is a necessary condition for Regular languages but not sufficient condition
i.e. All Regular languages must satisfy this condition and some Non-regular languages also satisfy this condition.
So, Regular Language → Satisfies Pumping Lemma
∴ Contrapositive statement is Doesn't Satisfy Pumping Lemma → Not Regular Languages
Pumping lemma for Regular languages says
" If a language L is Regular,
then $\exists P \geq 1$, such that
$\forall \,\,strings\,\,w \in L$, If $|w| \geq P$ then
$\exists x,y,z$, such that $w = xyz$
and $|xy| \leq P$
and $y \neq \in$ i.e. $|y| \geq 1$
and $\forall q \geq 0$, $xy^qz \in L$
In Words, If $L$ is regular, then there’s some magic number $P$(Called Pumping length). And if I take any string $w \in L$ that is at least as long as $P$, then I can break it up into three parts $x, y, $and $z.$ Now, the length of $xy$ is less than or equal to $P$, and also the length of $y$ is greater than or equal to $1$ and less than or equal to $P$.
In very simple words, If $L$ is a regular language then there is some positive number $P$ associated with $L$ such that for all strings $w \in L$ of length greater than or equal to $P$, we can find some non-empty sub-string $y$ in $w$ within the first $P$ symbols of $w$ such that when we repeat $y$ Zero or more times then the produced strings also belong to $L.$
Now, Coming to the language in the Question, this language consists of the following strings.
$L = \{ a^2, a^5,a^8,a^{11},a^{14},........., b^{10},b^{22},b^{34},b^{46}.......... \}$
Let me assume that Pumping length is $P = 10$ (Note that if Minimum pumping length for a language is $x$ then any number $\geq x$ is also a Pumping length for the language )
Since, I have assumed that Pumping length is $P = 10$, so, For all strings whose length is $\geq 10$ should be Pumped i.e. If $w $ is a string of length $\geq 10$ then within the first $P \,\,i.e. 10$ symbols of the string $w$, we should be able to find some non-empty substring $y$, such that we can repeat $y$ Zero or More times and the resulted string still belongs to language $L.$
So, since $w = b^{10}$ has length $\geq 10$ so, $b^{10}$ must be pumped. But in $b^{10} $ , If you take any non-empty substring and you remove that substring (Note that repeating substring $y$ Zero times in $w$ is equivalent to saying that remove $y$ from the string $w$) then resulted string does not belong to the given language. So, Pumping length cannot be 10. And Since we know that if Minimum pumping length for a language is $x$ then any number $\geq x$ is also a Pumping length for the language. So, Since 10 is Not a pumping length for $L$, so, any number $\leq 10$ cannot be a Pumping length for $L.$
So, Option 1,2,3 can be eliminated.
Moreover, The minimum pumping length for this language is $12.$ So, any number $\geq 12$ is a Pumping length for the given language.
Minimum Pumping length for the given language is P = 12 :
Take any string $w $ of length $\geq 12.$
Say, you take $a^{14}$ then the non-empty substring $y$ within the first $P\,\,i.e.12$ symbols can be taken as first three $a's$ i.e. $aaa.$ When you repeat $aaa$ Zero or more times, the resulted string still belongs to the language.
Similarly, for any $a^{2+3k}, k \geq 4$, $y$ can be chosen as the first three $a's$ i.e. $aaa.$ When you repeat $aaa$ Zero or more times, the resulted string still belongs to the language.
Say, you take $b^{22}, $ then the non-empty substring $y$ within the first $P\,\,i.e.12$ symbols can be taken as first twelve $b's$ i.e. $b^{12}.$ When you repeat $b^{12}$ Zero or more times, the resulted string still belongs to the language.
Similarly, for any $b^{10+12k}, k \geq 1$, $y$ can be chosen as the first twelve $b's$ i.e. $b^{12}.$ When you repeat $b^{12}$ Zero or more times, the resulted string still belongs to the language.
So, Minimum Pumping length for this given language $L$ is 12. And So, any number $\geq 12 $ is a Pumping length for L.
Extra Notes :
Coming to Minimum Pumping length(I'll use abbreviation MPL), MPL is the least possible value of $P$ such that Pumping lemma is satisfied by the regular language.
Some Facts :
I recommend to go through the Proof of Pumping lemma for Regular languages or at least the Idea of Pumping lemma based on Pigeon hole principle. Here : https://www.youtube.com/watch?v=vUHJW-OGvFE&list=PLIPZ2_p3RNHjGbysj9OvLTfL2qhsTdsbr&index=4
1. If you see the intuitive Proof of Pumping lemma(based on pigeon hole principle), You will find that $MPL \leq n$ where $n$ is the number of states in the minimal DFA accepting the regular language. So, If you draw mDFA(minimal DFA) for the given regular language and say the number of states in mDFA is $n$ then you can say that $MPL $ will be less than or equal to $n.$
2. MPL will always be strictly greater than the minimal string in the language. (Hint for Proof : Since $y$ can be repeated Zero or more times, When you repeat $y$ Zero times, the string length decreases by at least one symbol)
3. In the definition of Pumping lemma, $P \geq 1$ so, $MPL \geq 1$
4. If mDFA has a Dead state then $MPL \leq n-1$ (Hint for proof : If there is a Dead state, then all the strings accepted by the mDFA will be among the remaining $n-1$ states, so, if the language is Infinite, then the Loop must be between the remaining states.)
5. If language is Finite then MPL will be $x+1$ where $x$ is the length of the longest string in the language.
6. If Minimum pumping length for a language is $x$ then any number $\geq x$ is also a Pumping length for the language.
More examples to practice Minimum Pumping length : https://gateoverflow.in/303738/michael-sipser-exercise?show=303856#a303856
$\color{Red}{\text{Understand Complete Pumping Lemma, Crystal Clear:}}$
Here is Complete Pumping Lemma Playlist https://www.youtube.com/playlist?list=PLIPZ2_p3RNHjGbysj9OvLTfL2qhsTdsbr
This Pumping Lemma playlist contains EVERYTHING about Pumping Lemma of Regular Languages, i.e. Proof, Examples, Variations, GATE PYQs, Finding minimum pumping length etc. Watch this lecture for Complete, Correct & Clear Understanding.