Answer
Yes, $6a(a+b)$ is a multiple of $3a$. To see this, note that $6a(a+b)=3a[2(a+b)]$, where $3a$, $2$, and $a+b$ are all integers because they are sums and products of integers. By definition, then, $6a(a+b)$ is a multiple of $3a$.
Work Step by Step
Note that "B is a multiple of A" means the same thing as "A divides B".