how to find mixed Nash equilibria for 3x3? A (3,2)(3,0)(2,2)
B (1,0)(3,3)(0,3)
C (0,2)(0,0)(3,2)
p      q    1-p-q
So what I have done is : 
3p+3q+2(1-p-q)=p+3q
q=1
this is when A=B
p+3q=3(1-p-q)
p=-3/4
this is when B=c 
I don't know where i went wrong ,but how come p is negative? 
 A: You're on the right track. For there to exist a mixed equilibrium in which P1 mixes over $A$ and $B$, P1 must be indifferent between those two actions, i.e. his expected payoff must be the same from each. This implies conditions on P2's mix, which you denote by $p$ and $q$.
For P1 to be indifferent between $A$ and $B$, I get, as you do, that
$$3p+3q+2(1-p-q) = p+3q \Leftrightarrow q=1$$
For P1 to be indifferent between $B$ and $C$, I get, as you do, that
$$p+3q=3(1-p-q)$$
These two statements contradict (or imply $p$ is negative, which won't work), therefore there exists no mix of P2 actions such that P1 is indifferent between all three of his actions. That is, there is no mixed equilibrium in which P1 puts positive weight on $A$, $B$, and $C$.
However, there may be a mixed equilibrium where P1 puts weight on exactly two of his actions. So you have ruled out only one possibility so far. Now try to find mixed equilibrium with P1 playing an $AB$ mix, a $BC$ mix, an $AC$ mix, as well as when P1 is playing pure strategies - for instance, I can see already that there exists one mixed equilibrium where P1 plays just A, and P2 mixes over $a$ and $c$.
