A, V and Y alone can do a job in 6 weeks, 9 weeks and 12 weeks respectively. They work together for 2 weeks. Then A leaves the job. V leaves the job a week earlier to the completion of the work. The job would be completed in:

Amount of work done by A in one week = $$\frac{1}{6}$$

Amount of work done by V in one week = $$\frac{1}{9}$$

Amount of work done by Y in one week = $$\frac{1}{12}$$

Therefore Amount of work done by A, V and Y in one week = $$\frac{1}{6}$$ + $$\frac{1}{9}$$ + $$\frac{1}{12}$$ = $$\frac{13}{36}$$

They work together for 2 weeks therefore work done in these 2 weeks is = $$\frac{26}{36}$$

Amount of work remaining is = 1 - $$\frac{26}{36}$$ = $$\frac{10}{36}$$

Amount of work done by V and Y in one week = $$\frac{1}{9}$$ + $$\frac{1}{12}$$ = $$\frac{7}{36}$$

If total number of weeks required to complete the work is n then only V and Y work together for (n-3) weeks

Therefore work done in these (n-3) weeks is = $$\frac{7*(n-3)}{36}$$

Remaining $$\frac{10}{36}$$ - $$\frac{7*(n-3)}{36}$$ is completed by Y in one week

Therefore $$\frac{10}{36}$$ - $$\frac{7*(n-3)}{36}$$ = $$\frac{1}{12}$$

Solving we get n = 4

Therefore The job would be completed in 4 weeks 

Therefore our answer is option 'A'