This would be impractical with a single regular expression. You could get it to work with negative lookbehinds, but it would get quite complicated with each additional delimiter. It's pretty trivial to do this with plain old str.split() and multiple lines. All you have to do is check if splitting with the current delimiter gives you two elements. If it does, that's your answer. If not, move on to the next delimiter:

def split_new(inp, delims):
    for d in delims:
        result = inp.split(d, maxsplit=1)
        if len(result) == 2: return result

    return [inp] # If nothing worked, return the input

To test this:

teststrs = ['121 34 adsfd' , 'dsfsd and adfd', 'dsfsd & adfd' , 'dsfsd - adfd' , 'dsfsd and adfd and adsfa' , 'dsfsd and adfd - adsfa' , 'dsfsd - adfd and adsfa' ]
for t in teststrs:
    print(repr(t), '->', split_new(t, ['and', '&', '-']))

outputs

'121 34 adsfd' -> ['121 34 adsfd']
'dsfsd and adfd' -> ['dsfsd ', ' adfd']
'dsfsd & adfd' -> ['dsfsd ', ' adfd']
'dsfsd - adfd' -> ['dsfsd ', ' adfd']
'dsfsd and adfd and adsfa' -> ['dsfsd ', ' adfd and adsfa']
'dsfsd and adfd - adsfa' -> ['dsfsd ', ' adfd - adsfa']
'dsfsd - adfd and adsfa' -> ['dsfsd - adfd ', ' adsfa']

Try:

import re

tests = [
    ["121 34 adsfd", ["121 34 adsfd"]],
    ["dsfsd and adfd", ["dsfsd ", " adfd"]],
    ["dsfsd & adfd", ["dsfsd ", " adfd"]],
    ["dsfsd - adfd", ["dsfsd ", " adfd"]],
    ["dsfsd and adfd and adsfa", ["dsfsd ", " adfd and adsfa"]],
    ["dsfsd and adfd - adsfa", ["dsfsd ", " adfd - adsfa"]],
    ["dsfsd - adfd and adsfa", ["dsfsd - adfd ", " adsfa"]],
]

for s, result in tests:
    res = re.split(r"and|&(?!.*and)|-(?!.*and|.*&)", s, maxsplit=1)
    print(res)
    assert res == result

Prints:

['121 34 adsfd']
['dsfsd ', ' adfd']
['dsfsd ', ' adfd']
['dsfsd ', ' adfd']
['dsfsd ', ' adfd and adsfa']
['dsfsd ', ' adfd - adsfa']
['dsfsd - adfd ', ' adsfa']

Explanation:

The regex and|&(?!.*and)|-(?!.*and|.*&) uses 3 alternatives.

1. We match and always or:
2. We match & only if there isn't and ahead (using the negative look-ahead (?! ) or:
3. We match - only if there isn't and or & ahead.

We're using this pattern in re.sub -> splitting only on the first match.

You can keep a list of the delimiters, ordered by their value. Then, you can combine re.split with re.findall to only use delimiters produced from the latter that are the least valuable in the split, per the ranking in ops:

import re
def split_order(s):
   r, ops = re.findall('(?<=\s)and(?=\s)|\&|\-', s), ['and', '&', '-']
   m = -1 if not r else min([ops.index(i) for i in r])
   a, *b = re.split('|'.join(l:=[i for i in r if ops.index(i) == m]), s)
   return [s] if not l else ([a] if not b else [a, s[len(a)+len(l[0]):]])


vals = ['121 34 adsfd' , 'dsfsd and adfd', 'dsfsd & adfd' , 'dsfsd - adfd' , 'dsfsd and adfd and adsfa' , 'dsfsd and adfd - adsfa' , 'dsfsd - adfd and adsfa' ]
for i in vals:
   print(split_order(i))

Output:

['121 34 adsfd']
['dsfsd ', ' adfd']
['dsfsd ', ' adfd']
['dsfsd ', ' adfd']
['dsfsd ', ' adfd and adsfa']
['dsfsd ', ' adfd - adsfa']
['dsfsd - adfd ', ' adsfa']