Three continuous natural numbers whose product is 336

Three continuous natural numbers whose product is 336

336 = 3 X 112 = 3 X 2 X 56 = 3 X 2 X 7 X 8 = 6 X 7 X 8
These three consecutive natural numbers are 6, 7 and 8

For any natural number, sum all its digits, multiply the sum by 3 and add 1, repeat this operation, and finally get a fixed number is many

13
Such as 123
1+2+3=6
6*3+1=19
1+9=10
10*3+1=31
3+1=4
4*3+1=13
1+3=4
4 * 3 + 1 = 13 (cycle)
...

For any natural number, first sum all its digits, then multiply the sum by 3 and then add 1. Repeat this operation for many times, and the result is fixed
This fixed number is:
A.0 B.1 C.13 D.12
Please explain the process

Suppose that the number is one digit x, which obviously does not meet the requirements;
Suppose the number is two digits XY, according to the meaning of the problem, we can solve the equation 10 * x + y = 3 * (x + y) + 1, and get 7x = 2 * y + 1, obviously only x = 1, y = 3 is the answer;
Assuming that this number is 3-digit XYZ, it obviously does not meet the requirements
To sum up, whether there are options or not, we can get the answer