Divisibilty

Here, $12ab$ is divisible by $2,5$ . So the last digit $b$ must be $0$ .

Again, for divisibility by $7$ , here multiplying $b$ or $0$ by $2$ and then subtract from the rest, then which result comes must be divisible for $7$

Now, $12a-0=12a$

This $12a$ number must be divisible by $7$ and it is possible for $a=6$

Then the number comes $1260$

The result $a+b=6+0=\boxed6$