To find the common difference of an arithmetic progression (AP) where the 7th term is twice the 3rd term, and the first term is 12, follow these steps:

1. Recall the formula for the $n$-th term of an AP:

$$a_n=a+(n-1)\cdot d$$

where $a$ is the first term and $d$ is the common difference.

2. Write the expressions for the 7th term and the 3rd term:

- The 7th term ($a_7$) is:
$$a_7=a+6d$$

- The 3rd term ($a_3$) is:
$$a_3=a+2d$$

3. According to the problem, the 7th term is twice the 3rd term:

$$a+6d=2(a+2d)$$

4. Substitute the first term $a=12$ into the equation:

$$12+6d=2(12+2d)$$

5. Expand and simplify:

$$12+6d=24+4d$$
$$6d-4d=24-12$$
$$2d=12$$
$$d=\frac{12}{2}=6$$

Thus, the common difference $d$ is 6, which corresponds to option A.