According to the question,

In 1 day

work completed by (A+B)= $\frac{1}{72}$

work completed by (B+C)= $\frac{1}{120}$

work completed by (A+C)= $\frac{1}{90}$

2(A+B+C)'s  1day's work $=\frac{1}{72}+\frac{1}{120}+\frac{1}{90}$

∴ (A+B+C)'s 1 day's work $=\frac{5+3+4}{360}\times \frac{1}{2}=\frac{12}{360\times 2}=\frac{1}{60}$

∴ No. of days in which A alone will complete the work= (A+B+C)'s 1day work - (B+C)'s 1 day work

$=\frac{1}{60}-\frac{1}{120}=\frac{2-1}{120}=\frac{1}{120}$

∴ A alone can finish the work in 120 days

Hence option C is the correct.