Method 1:

class Solution:
    def minFlips(self, a: int, b: int, c: int) -> int:
        '''
            We need to change a and b
            Result : a | b = c
            | is a bitwise or
        '''
        # Find binary value for each
        binA, binB, binC  = bin(a)[2:], bin(b)[2:], bin(c)[2:]
        l1 = len(binA)
        l2 = len(binB)
        l3 = len(binC)

        # Find maximum length
        maxLen = max(l1, l2, l3)

        # Fill empty places with zeroes
        binA = binA.zfill(maxLen)
        binB = binB.zfill(maxLen)
        binC = binC.zfill(maxLen)
        ans = 0
        for i in range(maxLen):
            req = binC[i]
            if req == '1':
                if binA[i] == '0' and binB[i] == '0':
                    ans += 1 
            else:
                if binA[i] == '1' and binB[i] == '1':
                    ans += 2
                elif binA[i] == '1' or binB[i] == '1':
                    ans += 1
        return ans

Method 2 (Short):

class Solution:
    def minFlips(self, a: int, b: int, c: int) -> int:
        '''
        We need to find the LSB : Least significant bit i.e. Last binary value
        eg. for 100 : 0 is LSB
            for 101 : 1 is LSB

        We can use '& 1' to get the LSB

        For each Iteration we need to shift towards the right
        We can use right shift operator >>
        eg. 100 >> 1 would become 10
        '''
        ans = 0
        # Until one of them has value
        while a or b or c:
            # Find LSB (Last value)
            bitA = a & 1
            bitB = b & 1
            bitC = c & 1

            if bitC:
                ans += not (bitA  + bitB)
            else:
                ans += (bitA + bitB)

            # Right shift all of the values
            a >>= 1
            b >>= 1
            c >>= 1

        return ans