first balls: a b
second balls: c d
third balls: e f

now we weight, b and c.. d and e only..
if b = c, d = e then we can conclude b c f / a d e
if b =/c , d =/e then we can conclude b d f / a c e ( if b and d heavier )
if b = c, d =/e then b c e / a d f.

you dont have to know which one heavier or lighter, you can simply group them up with simple measurement
question?

27 coins = 8 8 8 1 1 1. thats a b c d e f
if a = b, a = c, we go to the ones.. if d = e, then the answer is F. COUNT = 3
if a = b, a = c, we go to the ones.. if d =/ e, we weight f. if d =/ f then the answer is D. COUNT = 4
if a = b, a = c, d =/ e, d = f, the answer is E. COUNT = 4

now we go to the if the diff coin is at a or b or c.
if a = b, a =/ c, then we measure c.

before we go further, i want to show you how we solve 8 coins with 3 scaling.
8 coins we devide it into twos.

8 coins = 2 2 2 2, thats w x y z.
if w = x, w = y, then we weight 1 coin from w, and 1 coin from z. if w = z, then the other coin from z is the answer.
if w = x, w =/ y, then we weight 1 coin from w, and 1 coin from y. if w =/ y, the answer is that coin from y.
if w =/ x, w = y, then we weight 1 coin from w, and 1 coin from x. if w = x, the other coin from x is the asnwer.
if w =/ x, w =/ y, one coin from W is the answer. we weight 1 coin from W, 1 coin from x or y or z..
TOTAL COUNT = 3

if a = b, then the coin is at C. total count 4.
if a =/ b, a =/ c, then the coin is at A. total count 5.
if a =/ b, a = c, then the coin is at B. total count 5.