Automata that recognizes Kleene closure of permutations of three symbols
There are a few things wrong with this automata:


It does not accept the empty string.
It is not a DFA – there should be no undefined transitions in a DFA.
It only accepts strings of length three, not multiples of three.


You are on the right track. Dealing with 1. and 3. above are fairly straightforward.
 Feed in an empty string. What state do you end up in? What does it mean to accept?
Feed in some longer strings. What's the pattern?
Dealing with 2. is rudimentary too, just a matter of filling in the missing transitions.

It's a good idea to read and understand again the definition of DFA, in particular, understanding what is allowed and what is not allowed.

EDIT: Here is a prettier solution: