On a fixed beam, which force induces a counterclockwise torque

Given this situation ($P$ is the point where the beam is fixed), which force induces a torque that is counterclockwise?

I want to use that : $ boldsymbol{tau} = textbf{r} times textbf{F}$ , however what does counter clockwise torque mean? Intuitionally it would seem that $F_2$ would turn the beam around its own axis (counter clockwise) but so would $F_3$ and $F_5$.

So my question actually is: what does counter clockwise torque mean in the context of this exercise?

Any hints are greatly appreciated.