Correct Answer: $\frac{120}{11}$

Solution : Time taken by A and B to do the work = 10 days
Let the work done by A, B, and C in a day be a, b and c units, respectively.
Work done by A and B together in a day = $\frac{1}{10}$
⇒ a + b = $\frac{1}{10}$ ------------(i)
Time taken by B and C to do the work = 12 days
Work done by B and C together in a day = $\frac{1}{12}$
⇒ b + c = $\frac{1}{12}$ ------------(ii)
Time taken by C and A to do the work = 6 days
Work done by C and A together in a day = $\frac{1}{6}$
⇒ c + a = $\frac{1}{6}$ ------------(iii)
Subtracting (ii) from (i),
⇒ a – c = $\frac{1}{10}$-$\frac{1}{12}$ = $\frac{2}{120}$
⇒ a – c = $\frac{1}{60}$ -------------(iv)
Adding equations (iii) and (iv),
⇒ 2a = $\frac{1}{6}$+$\frac{1}{60}$ = $\frac{11}{60}$
$\therefore$ a = $\frac{11}{120}$
Work done by A alone in a day = $\frac{11}{120}$
$\therefore$ Time taken by A alone to do the work = $\frac{120}{11}$ days
Hence, the correct answer is $\frac{120}{11}$.