How to compare multi column values with other multi column value of same dataframe?

You can get all permutations of b columns, left join with original DataFrame filtered only rows with missing values in a3 for a3_ column wich match a1, a2 in list. Then join list to one Series, remove possible duplicates in index and replace missing values of a3 column in original DataFrame:

from  itertools import permutations

cols = ['b1','b2','b3']
L = [df[df['a3'].isna()].merge(df.loc[:, x].set_axis(['a1','a2','a3'], axis=1),
              how='left', on=['a1','a2'], suffixes=('','_'))['a3_'].dropna()
     for x in permutations(cols, 3)]

final = pd.concat(L)
final = final[~final.index.duplicated()]
print (final)
4    66.0
1    89.0
Name: a3_, dtype: float64

df['a3'] = df['a3'].fillna(final)
print (df)
          time  duration  a1  a2    a3  b1  b2  b3
0  2022-02-28         95  11   2   3.0  22  67  25
1  2022-02-27         85  84   5  89.0  72  23  15
2  2022-02-26         87   6   7   8.0   5  84  89
3  2022-02-25         72   9  10  44.0  55  78  41
4  2022-02-24         66  19  57  66.0  50  60  51
5  2022-02-23         88  20  48  67.0  19  66  57

I use to code algorithm by hand trying to be readable and understandable.
Here is an old geek fashion code with the following algoritm:

if a1,a2 are in b1,b2,b3 respective products are a1a2 and a1a2the third element.
the thirdElement is exactly b1b2b3//a1a2 if the remainder is 0. (notice integer division).
As a1a2 can be equal to several bxby , check too that a1+a2 == bx+by (where bx+by is (b1+b2+b3 - third element) ).
(a3 is set to None here, could be "" depending of your data )

tests = [ \
["2022-02-28",95,11,2,3,22,67,25],\
["2022-02-27",85,84,5,None,72,23,15],\
["2022-02-26",87,6,7,8,5,84,89],\
["2022-02-25",72,9,10,44,55,78,41],\
["2022-02-24",66,19,57,None ,50,60,51],\
["2022-02-23",88,20,48,67,19,66,57]\
]
# first collect products and sums of all b
storeb1b2b3 =[]
storesumb1b2b3 = []
for test in tests:
    storeb1b2b3.append(test[5]*test[6]*test[7])
    storesumb1b2b3.append(test[5]+test[6]+test[7])

# now loop on a1 a2 and check in all b 
for ia in range(0,len(tests)):
    test = tests[ia] 
    # check a3 empty
    if (test[4] != None):
        continue
    a1= test[2]
    a2 = test[3]
    a1a2 =a1*a2
    # check against all b 
    for ib in range(0,len(storeb1b2b3)):
        b1b2b3= storeb1b2b3[ib]
        sumb1b2b3 = storesumb1b2b3[ib]
        if (b1b2b3 % a1a2 == 0):
            candidate =(b1b2b3//a1a2)
            if ((sumb1b2b3 - candidate) == (a1+a2)):
                print ( "for",a1,a2,"at",tests[ia][0],"found: ",candidate, "at ",tests[ib])
    

Notes :
products and sum of b are calculated once at begining to optimize. If a too long list to be stored, can be calculated on the fly in a1a2 loops.
Use indices ia, ib to be able to print source lines.

for 84 5 at 2022-02-27 found:  89 at  ['2022-02-26', 87, 6, 7, 8, 5, 84, 89]
for 19 57 at 2022-02-24 found:  66 at  ['2022-02-23', 88, 20, 48, 67, 19, 66, 57]

Hope you enjoy coding too :)

addendum : someone asks me to prove that if ab=a'b' and a+b=a'+b' then pair [a,b] is unique. Demonstration.

ab = a'b' and a+b = a'+b'
if a = a' , as (a+b = a'+b') => b = b' : they are same numbers.
if a != a' => a' = a + x, as (a+b = a'+b') => b' = b - x
products are :
ab = (a+x)*(b-x) = ab - xa + xb -x2 = ab +x(-a+b-x).
x(-a+b-x) = 0.
two cases :
1) x = 0 so a'= a+0 and b' = b-0 still same values a,b.
2) -a+b-x = 0
x = b-a
so a'= a+x = b.
b'= b-x = a
just a permutation of a,b in b,a => in any case if (ab = a'b')and (a+b=a'+b') this is always the same pair [a,b].

QED