• How have the forces be prepared, in order to add them correctly?

\begin{align*} F_0 &= |\vec{F_0}| \quad \quad \text{ with } \vec{F_0} = \left( \begin{matrix}{F_{x,0}}\\ {F_{y,0}} \end{matrix} \right) = \left( \begin{matrix} \sum\limits_{n} F_{x,0n} \\ \sum\limits_{n} F_{y,0n} \end{matrix} \right) \\ F_0 &= \sqrt{ \left(\sum\limits_{n} F_{x,0n} \right)^2 + \left(\sum\limits_{n} F_{y,0n} \right)^2 } \\ \end{align*}

The forces have to be resolved into coordinates. Here, it is recommended to use an orthogonal coordinate system ($x$ and $y$).
The coordinate system shall be in such a way, that the origin lays in $Q_0$, the x-axis is directed towards $Q_3$ and the y-axis is orthogonal to it.
For the resolution of the coordinates it is necessary to get the angles $\alpha_{0n}$ of the forces with respect to the x-axis.
In the chosen coordinate system this leads to: $\alpha_{0n} = atan(\frac{\Delta y}{\Delta x})$
$\alpha_{01} = \rm{atan}(\frac{3}{1})= 1.249 = 71.6°$
$\alpha_{02} = \rm{atan}(\frac{4}{3})= 0.927 = 53.1°$
$\alpha_{03} = \rm{atan}(\frac{0}{3})= 0= 0°$

Consequently, the resolved forces are:

\begin{align*} F_{x,0} &= F_{x,01} + F_{x,02} + F_{x,03} && | \quad \text{with } F_{x,0n} = F_{0n} \cdot \rm{sin}(\alpha_{0n}) \\ F_{x,0} &= (-5~\rm{N}) \cdot \rm{sin}(71.6°) + (-6~\rm{N}) \cdot \rm{sin}(53.1°) + (+3~\rm{N}) \cdot \rm{sin}(0°) \\ F_{x,0} &= -2.18 ~\rm{N} \\ \\ F_{y,0} &= F_{x,01} + F_{x,02} + F_{x,03} && | \quad \text{with } F_{y,0n} = F_{0n} \cdot \rm{cos}(\alpha_{0n}) \\ F_{y,0} &= (-5~\rm{N}) \cdot \rm{cos}(71.6°) + (-6~\rm{N}) \cdot \rm{cos}(53.1°) + (+3~\rm{N}) \cdot cos(0°) \\ F_{y,0} &= -9.54 ~\rm{N} \\ \\ \end{align*}

\begin{align*} F_0 &= \sqrt{ (-2.18 ~\rm{N})^2 + (-9.54 ~\rm{N})^2 } = 9.79 ~\rm{N} \rightarrow 9.8 ~\rm{N} \\ \end{align*}