Let the total capacity of the tank be 1 unit (i.e., the tank is completely filled when the total capacity is reached).
Rate of filling:
- Tap A fills the tank in 12 hours, so the rate of tap A is \( \frac{1}{12} \) of the tank per hour.
- Tap B fills the tank in 15 hours, so the rate of tap B is \( \frac{1}{15} \) of the tank per hour.
- Tap C fills the tank in 20 hours, so the rate of tap C is \( \frac{1}{20} \) of the tank per hour.
Filling pattern:
- Tap A is open all the time.
- Tap B and Tap C are opened alternately for 1 hour each.
In 2 hours:
- In the first hour, tap A and tap B are open. The amount of water filled is:
\[
\text{Water filled in 1st hour} = \frac{1}{12} + \frac{1}{15} = \frac{5}{60} + \frac{4}{60} = \frac{9}{60} = \frac{3}{20}
\]
- In the second hour, tap A and tap C are open. The amount of water filled is:
\[
\text{Water filled in 2nd hour} = \frac{1}{12} + \frac{1}{20} = \frac{5}{60} + \frac{3}{60} = \frac{8}{60} = \frac{2}{15}
\]
So, in 2 hours, the total water filled is:
\[
\frac{3}{20} + \frac{2}{15} = \frac{9}{60} + \frac{8}{60} = \frac{17}{60}
\]
Total time to fill the tank:
Now, we need to calculate how many such 2-hour cycles are needed to fill the tank.
Let’s calculate how many cycles are needed to fill the tank. The total amount of water filled after \( x \) cycles is:
\[
\frac{17}{60} \times x = 1
\]
Solving for \( x \):
\[
x = \frac{60}{17} \approx 3.53 \quad \text{(approximately 3 full cycles and part of the 4th cycle)}
\]
Since each cycle takes 2 hours, the total time taken to fill the tank is approximately:
\[
2 \times 3.53 = 7.06 \, \text{hours}
\]
So, the tank will be filled in approximately 7 hours.
Thus, the correct answer is (1) 7 hours.