A and B can complete a piece of work in 12 days. B and C can complete the same work in 15 days and C and A in 10 days. How many days will A take to complete the work by himself?

The correct option is $17\frac{1}{7}$.

Let the total work to be completed be represented by 1 unit.

From the given problem, we can find the one-day work (work rate) of each pair:

1. A and B together complete the work in 12 days. Thus, their 1-day work is:

$(A+B)\text{'s 1-day work}=\frac{1}{12}$

2. B and C together complete the work in 15 days. Thus, their 1-day work is:

$(B+C)\text{'s 1-day work}=\frac{1}{15}$

3. C and A together complete the work in 10 days. Thus, their 1-day work is:

$(C+A)\text{'s 1-day work}=\frac{1}{10}$

Now, by adding the 1-day work of all three pairs together, we get twice the combined 1-day work of A, B, and C:

$2(A+B+C)\text{'s 1-day work}=\frac{1}{12}+\frac{1}{15}+\frac{1}{10}$

To add these fractions, we find the Least Common Multiple (LCM) of 12, 15, and 10, which is 60:

$2(A+B+C)\text{'s 1-day work}=\frac{5+4+6}{60}=\frac{15}{60}=\frac{1}{4}$

Dividing by 2 gives the 1-day work of A, B, and C working together:

$(A+B+C)\text{'s 1-day work}=\frac{1}{4\times 2}=\frac{1}{8}$

To find the 1-day work of A alone, subtract the 1-day work of (B + C) from the combined 1-day work of (A + B + C):

$A\text{'s 1-day work}=\frac{1}{8}-\frac{1}{15}$

The LCM of 8 and 15 is 120:

$A\text{'s 1-day work}=\frac{15-8}{120}=\frac{7}{120}$

The total time taken by A to complete the work alone is the reciprocal of A's 1-day work:

$\text{Time taken by A alone}=\frac{120}{7}\text{days}$

Converting $\frac{120}{7}$ into a mixed fraction:

$120÷7=17\text{with a remainder of}1$

$\text{Time taken by A}=17\frac{1}{7}\text{days}$

Therefore, A alone can complete the work in $17\frac{1}{7}$ days.