A and B together can do a work in 10 days. B and C together can do the same work in 6 days. A and C together can do the work in 12 days. Then A, B and C together can do the work in

A and B togeather complete 1/10 th of work in a day.
$$(\frac{1}{A} + \frac{1}{B}) = \frac{1}{10}$$ --------------- 1
B and C togeather complete 1/6 th of work in a day.
$$(\frac{1}{B} + \frac{1}{C}) = \frac{1}{6}$$ ----------------- 2
A and C togeather complete 1/12 th of work in a day.
$$(\frac{1}{A} + \frac{1}{C}) = \frac{1}{12}$$ --------------- 3
Adding 1 ,2 and 3
$$2 \times(\frac{1}{A} + \frac{1}{B} + \frac{1}{C}) = \frac{1}{10} + \frac{1}{6} + \frac{1}{12}$$
Taking LCM we get
$$\frac{1}{10} + \frac{1}{6} + \frac{1}{12} = \frac{6+10+5}{60} = \frac{21}{60} $$
$$(\frac{1}{A} + \frac{1}{B} + \frac{1}{C})= \frac{21}{120}$$
A B and C complete 21/120 th of the work in a day.
In 5 days they would complete 105/120th of the work.
At the end of 6th day they would have completed 126/120 th of work (Which is greater than 1)
Hence they finish the entire work before the end of 6th day.
Out of the given options Option C suits the best.
Hence Option C is the correct answer.