Since we need to consider the least portion of the tank to be filled by any of the three pipes, we will consider the three pipes $B, C$, and $D$, which are less efficient in filling.

Therefore, the least portion that can be filled by these three pipes in one day is $\(\frac{1}{6}+\frac{1}{9}+\frac{1}{12}=\frac{6+4+3}{36}=\frac{13}{36}\)$. Thus, the correct answer is $\(\frac{13}{36}\)$.