Simplest thing might be to use an affine transformation so that the minimum of each is $0$, the maximum is $100$, and the relative proportions stay the same when the minimum is and maximum are different, and all equal to the midpoint, 50, if all three entries are equal; so that, for example, with $\{0,50,75\}$ you want the middle number to be two thirds of the way between $0$ and $100$.
Say your set contains $a\leq b\leq c$. If $a=b=c$, then normalize it by transforming it into $(50,50,50)$.
If $a\lt c$, then replace each element $x$ with $100\left(\frac{x-a}{c-a}\right)$.
Doing this, your three sets would become $\{0, \frac{20}{3},100\}$, $\{0, 50, 100\}$, and $\{0, \frac{200}{3}, 100\}$.
In other words, given a multiset $S=\{a,b,c\}$, test to see whether $\max(S)=\min(S)$; if so, replace it with the multiset $\{50,50,50\}$. If $\max(S)\neq \min(S)$, then apply the function $x\longmapsto \frac{100(x-\min(S))}{\max(S)-\min(S)}$ to the multiset.
You can choose some other way to deal with the case where $a=b=c$, depending on what you want this for.