If A and B together can complete a piece of work in 20 days, B and C in 10 days and C and A in 12 days, then A, B and C together can complete the same work in :

$latex \mathbf{8\frac{4}{7}\ days}&s=1$
(A + B)’s 1 day’s work $latex = \frac{1}{20}&s=1$
(B + C)’s 1 day’s work $latex = \frac{1}{10}&s=1$
(C + A)’s 1 day’s work $latex = \frac{1}{12}&s=1$
On adding,
2(A + B + C)’s 1 day’s work $latex = \frac{1}{20}+\frac{1}{10}+\frac{1}{12} = \frac{3+6+5}{60} = \frac{7}{30}&s=1$
∴ (A + B + C)’s 1 day’s work $latex = \frac{7}{60}&s=1$
∴ Hence, the work will be completed in $latex \frac{60}{7} = 8\frac{4}{7}\ days&s=1$.
Hence option [C] is correct answer.