The total number of 5-digit numbers, formed by using the digits $$1, 2, 3, 5, 6, 7$$ without repetition, which are multiple of $$6$$, is

We need to find the total number of 5-digit numbers formed using digits $$\{1, 2, 3, 5, 6, 7\}$$ without repetition that are multiples of 6.

A number is divisible by 6 if and only if it is divisible by both 2 and 3. For divisibility by 2 the last digit must be even (2 or 6), and for divisibility by 3 the sum of the digits must be divisible by 3.

We must choose 5 digits out of the 6. The sum of all 6 digits is $$1 + 2 + 3 + 5 + 6 + 7 = 24.$$ If we omit a digit $$d$$, the sum of the remaining digits is $$24 - d$$, and for divisibility by 3 we require $$24 - d \equiv 0 \pmod{3},$$ i.e., $$d \equiv 0 \pmod{3}.$$ From the available digits, this means $$d = 3$$ or $$d = 6.$$

Omitting 3 leaves the digits $$\{1, 2, 5, 6, 7\}$$ with sum $$21$$, which is divisible by 3. For divisibility by 2 the last digit can be 2 or 6 (2 choices), and the remaining 4 positions can be filled by the other 4 digits in $$4! = 24$$ ways, giving a count of $$2 \times 24 = 48.$$

Omitting 6 leaves the digits $$\{1, 2, 3, 5, 7\}$$ with sum $$18$$, which is divisible by 3. The only available even digit is 2 (1 choice for the last digit), and the remaining 4 positions can be filled by the other 4 digits in $$4! = 24$$ ways, giving a count of $$1 \times 24 = 24.$$

Combining these cases yields a total of $$48 + 24 = 72,$$ so the correct answer is Option A: $$72$$.