In the given problem in 1 day,

(A+B) can do $\frac{1}{72}$ part of work

(B+C) can do $\frac{1}{120}$ part of work

(C+A) can do $\frac{1}{90}$ part of work,

So, in one day, 2(A+B+C) can do,

$$\begin{gathered} 2(\mathrm{A}+\mathrm{B}+\mathrm{C})=\frac{1}{72}+\frac{1}{120}+\frac{1}{90} \\ \mathrm{A}+\mathrm{B}+\mathrm{C}=\frac{5+3+4}{360\times 2}=\frac{12}{360\times 2}=\frac{1}{60} \end{gathered}$$

1 day work of A=[1 day work of (A+B+C)]-[1 day work of (B+C)]

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

Hence, A alone can finish te work in 120 days

Hence option C is correct