What is Pumping length for Union of Regular languages?
A pumping length is a positive integer $p$ such that any string of $L$ of length at least $p$ can be decomposed in the form $xyz$ with $|xy|\leq p$, $|y|&gt;0$ and for all $n\geq0$ the words $xy^nz$ are in $L$.
Assume that $p&lt;12$. Then the word $b^{10+12}$ should be able to be decomposed as above. We cannot take $y$ with $1\leq|y|\leq p &lt;12$. Otherwise $xz=xy^0z$ wouldn't be in $L$.
We can take $p=12$ or larger.
If $10+12k=|b^{10+12k}|\geq p\geq 12$ then $k\geq1$ and we can decompose $b^{10+12k}$ as $xyz$, with $x=\epsilon$, $y=b^{12}$ and $z=b^{10+12(k-1)}$. Then $|xy|=12\leq p$ and for all $n\geq0$ we have $xy^nz=b^0b^{12n}b^{10+12(k-1)}=b^{10+12(n+k-1)}$ in $L$, since $n+k-1\geq k-1\geq0$
If $2+3k=|a^{2+3k}|\geq p\geq 12$, then $k\geq4$. We can decompose $a^{2+3k}$ as $xyz$ with $x=\epsilon$, $y=a^3$ and $z=a^{2+3(k-1)}$. We have that $|xy|=3\leq p=12$, $|y|=3\geq1$ and for all $n\geq0$ we have $xy^nz=a^0a^{3n}a^{2+3(k-1)}=a^{2+3(n+k-1)}$ in $L$, since $n+k-1\geq k-1\geq0$.
So, the minimum pumping length of $L$ seems to be $12$. Any larger value would also work in this case.