We know that:
\[
A = 2B \quad \text{and} \quad B = 2C
\]
So, \( A = 2B = 2 \times 2C = 4C \).
The average of the reciprocals of A, B, and C is:
\[
\frac{1}{A} + \frac{1}{B} + \frac{1}{C} = \frac{7}{12}
\]
Substitute \( A = 4C \) and \( B = 2C \) into the equation:
\[
\frac{1}{4C} + \frac{1}{2C} + \frac{1}{C} = \frac{7}{12}
\]
Taking the LCM:
\[
\frac{1}{4C} + \frac{1}{2C} + \frac{1}{C} = \frac{1 + 2 + 4}{4C} = \frac{7}{4C}
\]
Now, equating this to \( \frac{7}{12} \):
\[
\frac{7}{4C} = \frac{7}{12}
\]
Cross-multiply:
\[
4C = 12
\]
So, \( C = 3 \), and hence \( A = 4C = 4 \times 3 = 12 \).
Thus, the correct answer is \( A = 4 \).