Total numbers of 3 -digit numbers that are divisible by 6 and can be formed by using the digits 1 ,   2…

A number is divisible by $6$ when it is divisible by $2$ and $3$ both.

The three digit number can be $\underset{¯}{a}\underset{¯}{b}\underset{¯}{2}$.

$a+b$ can be $4,7,10$

If $a+b=4$ then $(a, b)$ can be $(1,3)$, $(3,1)$, $(2,2)$

If $a+b=7$ then $(a, b)$ can be$(2,5),(5,2),(3,4),(4,3)$ 

If $a+b=10$ then $(a, b)$ can be $(5,5)$

So, $8$ such cases when $2$ is at unit's place

Similarly, there exist $8$ such cases when $4$ is at unit's place.

Total $=16$ cases.

Hence this is the required answer.