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.