This is the correct complete solution:
Pipe A can fill a tank in 6 hours,
So Pipe A's tank filling rate is 1 tank per 6 hr or  or
 or 

pipe B can fill it in 2 hours less time than it takes drainpipe C to empty the tank (with pipes A and B closed). 
Suppose drainpipe C can drain 1 tank in x hours.  Then drainpipe C's "filling"
rate is a negative quantity -1 tank per x hours or  or
 or 
 
pipe B can fill it in 2 hours less time than it takes drainpipe C to empty the tank. 
So Pipe B's tank filling rate is 1 tank per x-2 hr or  or
 or 
 
With all three pipes open, it takes 3 hours and 20 minutes to fill the tank.
3 hours and 20 minutes is  or
 or  hours.  
So A, B, and C's combined filling rate is 1 tank per
 hours.  
So A, B, and C's combined filling rate is 1 tank per  hr or
 hr or  or
 or 
 or
 or 
 So we have this equation:
So we have this equation:






 
   





 
   





 Multiply through by LCD = 30x(x-2)
   5x(x-2) + 30x - 30(x-2) = 3·3x(x-2)
5x² - 10x + 30x - 30x + 60 = 9x(x-2)
            5x² - 10x + 60 = 9x² - 18x
                         0 = 4x² - 8x - 60
Divide through by 4      0 = x² - 2x - 15
                         0 = (x - 5)(x + 3)
                             x-5=0;   x+3=0
                               x=5;     x=-3 (ignore)
Aswer: 5 hours
Edwin
Multiply through by LCD = 30x(x-2)
   5x(x-2) + 30x - 30(x-2) = 3·3x(x-2)
5x² - 10x + 30x - 30x + 60 = 9x(x-2)
            5x² - 10x + 60 = 9x² - 18x
                         0 = 4x² - 8x - 60
Divide through by 4      0 = x² - 2x - 15
                         0 = (x - 5)(x + 3)
                             x-5=0;   x+3=0
                               x=5;     x=-3 (ignore)
Aswer: 5 hours
Edwin