The basic approach is to calculate the formula for expected value and maximize it by any tools you know (one usually uses linear programming).
If your adversary would play 'scissors' only, there is very specific pure optimal strategy (the mixed one would not be optimal).
In rock-paper-scissors a pure strategy would be rarely optimal, against unknown enemy the only optimal strategy is mixed $\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$.
It does not need to be unique, consider a game with payoff matrix being just zero everywhere (I know this is a silly game, still it is a zero-sum game, right?). Any strategy would be optimal there ;-)
Hope it helps ;-)
The procedure for finding mixed-strategy nash equilibrium should not be different when there are three players than when there are 2.
As in the two players' case, the key point is that if it is optimal for you to randomize between different actions, the expected payoff of each action must be the same (assuming that agents are expected utility maximizers). For example, if you randomize over two actions, say Rock and Paper, but ${U}((1,0,0), s_{-i}) > U((0,0,1), s_{-i})$ then you are definitely not optimizing.
Let $p_i(s)$ be the probability that player $i = 1,2,3$ plays action $s = r,p,s$. To get a mixed-strategy nash equilibrium,
$U_1((1,0,0),s_{-i}) = 0*p_2(r)*p_2(r) + (-1)*p_2(r)*p_3(p) + ...$
must be be equal to
$U_1((0,1,0),s_{-i}) = 2*p_2(r)*p_3(r) + (0.5)*p_2(r)*p_3(p) + ...$
which must itself be equal to
$U_1((0,0,1),s_{-i}) = (-1)*p_2(r)*p_3(r) + (0)*p_2(r)*p_3(p) + ...$
This should be true for every $i=1,2,3$ which leaves you with a simple system of equations to solve. A profile of strategy is a mixed-strategy Nash equilibrium if and only if it solves this system.
Based on that, it should not be too hard to determine whether there are other mixed-strategy nash equilibrium.
Best Answer
The expected payoff for $P1$ when he plays the mixed strategy $x$ and his opponent plays the mixed strategy $y$ is $x A y^\top$.