A cistern has two inlets A and B which can fill it in 12 hours and 15 hours respectively.

Time taken by tap A to fill the cistern $=12$ hours

Time taken by tap B to fill the cistern $=15$ hours

Let $C$ be the outlet that can empty the cistern in 10 hours.

Time taken by tap $\mathrm{C}$ to empty the cistern $=10$ hours

Now,

Tap A fills $\frac{1}{12}$ th part of the cistern in 1 hour.

Tap B fills $\frac{1}{15}$ th part of the cistern in 1 hour.

Tap C empties out $\frac{1}{10}$ th part of the cistern in 1 hour.

Thus, in one hour, $\left(\frac{1}{12}+\frac{1}{15}-\frac{1}{10}\right)$ th part of the cistern is filled.

We have:

$\frac{1}{12}+\frac{1}{15}-\frac{1}{10}=\frac{10+8-12}{120}=\frac{6}{120}=\frac{1}{20}$

Thus, in 1 hour, $\frac{1}{20}$ th part of the cistern is filled.

Hence, the cistern will be filled completely in 20 hours if all the three taps are opened together.