In a hostel there are 250 students, 120 watch Fox News, 80 watch Sky S

One formula for 3 overlapping sets is as follows:

Total = Group 1 + Group 2 + Group 3 - (at least 2 of the groups) + (all 3 groups) + none.

In the problem above:
Total = 250.
Group 1 = 120.
Group 2 = 80.
Group 3 = 90.
(at least 2 of the groups) = 50+60+65 = 175.
Let x = all 3.
Let N = none.

Plugging these values into the equation above, we get:
250 = 120 + 80 + 90 - 175 + x + N.
250 = 115 + x + N
N = 250 - 115 - x.
N = 135 - x.

To MINIMIZE the number who watch at least 1 channel, we must MAXIMIZE the value of N: the number who watch NONE of the channels.
In the equation above, the value of N will be maximized if x (all 3) is as small as possible.

Of the given overlaps -- 50, 60, 65 -- the two greatest overlaps both include A:
(at least S and A) + (at least F and A) = 60+65 = 125.
Now this value ((at least S and A) + (at least F and A)) is 35 greater than the total number who watch A (90).
Implication:
At least 35 students must watch all 3 channels.
Thus, the minimum value of x = 35.

Plugging x=35 into N = 135 - x, we get:
N = 135 - 35 = 100.
Since the maximum number who watch none of the channels = 100, the minimum number who watch at least 1 channel = 250-100 = 150.