The 7-digit numbers 7−4−a−5−2−b−1− and 3−2−6−a−b−4−c− are each multiples of 3. which of the following could be the value of c?

Since, you forgot the options, I can help only a little.

Multiples of 3 have the property that the sum of their digits has to be divisible by 3.

so for the first number it means that 7+4+A+5+2+B+1 must be divisible by 3.

let's simplify this:

7+4+A+5+2+B+1=11+A+7+B+1=19+A+B

we can actually from now on think in terms of :A+B+1 is divisibe by 3, since 19 =18+1 and 19
and the same for the second number
3+2+6+A+B+4+C=15+A+B+C

15 is divisible by 3, so we can just remember that

A+B+C  is divisible by 3.
A+B+1 is divisive by 3

C must be of the form 3x+2 then (since A+B+1 is divisible by 3_: it can be either 2 (3*0+2), 5, or 8 - since the next possibility, 11 is bigger than 10, so it cannot be a digit in this number!