jmc

Since both numbers are divisible by 3, the sum of their digits has to be divisible by three. $7+4+5+2+1=19$. To be a multiple of $3$, $A+B$ has to be either $2$ or $5$ or $8$... and so on. We add up the numerical digits in the second number; $3+2+6+4=15$. We then add two of the selected values, $5$ to $15$, to get $20$. We then see that C = $1,4$ or $7,10$... and so on, otherwise the number will not be divisible by three. We then add $8$ to $15$, to get $23$, which shows us that C = $1$ or $4$ or $7$... and so on. To be a multiple of three, we select a few of the common numbers we got from both these equations, which could be $1,4,$ and $7$. However, in the answer choices, there is no $7$ or $4$ or anything greater than $7$, but there is a $1$, so $\boxed{1}$ is our answer.