To be divisible by 3, the sum of the digits of a number must also be divisible by 3. The sum of the digits of $3A6,\!792$ is $3 + A + 6 + 7 + 9 + 2 = 27 + A$. The largest possible value of $A$ is 9, so let's see if that works. If $A = 9$, then the sum of the digits is $27 + 9 = 36$, which is divisible by 3. So the largest possible value of $A$ that makes the number divisible by 3 is 9. 
In the six-digit integer $3A6,\!792$, what is the largest digit $A$ so that the six-digit integer will be divisible by 3?