Answer : Option D

Explanation :

Note: Assume the total work = LCM of the given days

Take the LCM of days = LCM of (10, 15, and 30) = 30
Let the total work = 30

Note: One day work = (total work/ days)

Now,

A's one day work = 30/10 = 3 unit
B's one day work = 30/15 = 2 unit
C's one day work = 30/30= 1 unit

ATQ, A work continuously and B and C works alternatively

i.e., (A+B)'s one day work = 3+2 = 5 unit
And (A+C)'s one day work = 3+1 = 4 unit

Or, 2 days work = 9 unit
To complete the 27 units work, multiply both sides with 3.

i.e., 2 days * 3 = 9 unit * 3
Or, 6 days = 27 units of work
Now, on the 7th day, the remaining work is done by A + B.
If (A+B)'s one day work = 5 unit work
i.e.,
1 work = 1/5 days
Or, 3 work = 3/5 days

Hence, the work will be finished in 6+3/5 days or 6[3/5] days.