I have the following system (at P and Q there is a ball joint; the length of L beam is l; the PB and PA forces are applied at l/2 and are known; Fz is not zero and is known):
The system has 0 degrees of freedom because each ball joint removes three degrees of freedom (6-2*3=0). I replaced each ball joint with three unknown forces.
The problem is that the second cardinal equation along x gives:
$$-F_z \, l = 0$$
How can I solve the problem?
Thank you very much for your time.
The system is still statically indeterminate, but not due to lack of DOF. You have a singularity. Having correct DOFs only assures you have a chance of being statically determinate, but it does not guarantee this condition. In this case, the singularity is because the beam is allowed to rotate about the x axis unless $F_z$ is an unknown reaction.
As such, in this case, $F_zl = 0$ means $F_z = 0$. Here are your other equations, in no particular order:
$$\begin{alignat}{4} \sum& F_x &&= A + PB_x + D + PA_x &&= 0 \\ \sum& F_y &&= B + PB_y + E + PA_y &&= 0 \\ \sum& F_z &&= C + G + (F_z = 0) &&= 0 \\ \sum& M_{y@Q} &&= Cl &&= 0 \\ \sum& M_{x@Q} &&= F_zl &&= 0 \\ \sum& M_{z@Q} &&= -Bl + PB_yl/2 - PA_xl/2 &&= 0 \end{alignat}$$
We can immediately see $C = F_z = 0$. Because of $\sum F_z$, $G = 0$. We now have four unknowns and three equations. We can solve for $B$ immediately using:
$$\sum M_{z@Q} = -Bl + PB_yl/2 - PA_xl/2 = 0$$
Rearranging, we can use this result for $B$ to solve for $E$ using:
$$\sum F_y = B + PB_y + E + PA_y = 0$$
But the final equation is singular:
$$\sum F_x = A + PB_x + D + PA_x = 0$$
This cannot be resolved using statics, but require static indeterminate methods.
