ok, suppose the three subgroups are H,K,M.
we know there is x,y,z in G with:
xHx
-1 = K
yKy
-1= M
zMz
-1= H.
so the action is transitive. because we have an
action (these are bijective maps), if:
x

-->K
then xKx
-1 can only be H or M. if xKx
-1 = H, then x corresponds to the permutation (1 2).
otherwise xKx
-1 = M, and x corresponds to (1 2 3).
now if xKx
-1 corresponds to (1 2), we can look at y.
since y takes K to M, either y takes H to K, in which case y corresponds to (1 2 3)
or y takes H to H, in which case y corresponds to (2 3).
in this last case, we have xy corresponds to (1 2)(2 3) = (1 2 3):
xyH(xy)
-1 = x(yHy
-1)x
-1 = xHx
-1 = K
xyK(xy)
-1 = x(yKy
-1)x
-1 = xMx
-1 = M
xyM(xy)
-1 = x(yMy
-1)x
-1 = xKx
-1 = H.
(and we didn't even need to use z).