As @mg4w already mentioned in a comment, this is a statically indeterminate structure. This means that the structure can't be trivially solved as you have tried (unless you use a simplifying assumption, as I describe towards the end).
To show the problem, these are global equilibrium equations we have at our disposal:
$$\begin{align}
\sum F_x &= H_{top} + H_{mid} + H_{bot} = 0 \\
\sum F_y &= V_{top} + V_{mid} + V_{bot} - 10 = 0 \\
\sum M_{mid} &= -3.00H_{top} + 3.00H_{bot} - 3.00\times10 = 0 \\
&= -H_{top} + H_{bot} + 10 = 0 \\
\therefore H_{top} &= H_{bot} + 10
\end{align}$$
but this is as far as we can go. I could substitute $H_{bot}$ in the $\sum F_x = 0$, but that wouldn't get me anywhere. We're stuck. That's the trouble with hyperstatic (statically indeterminate) structures. Global equilibrium won't get you anywhere.
Now, it can be solved by hand, but it takes a bit of work. Here's your structure:
Let's start by simplifying your structure as far as we can to keep things clean once we reach the point where trivial methods will get stuck. For a start, observe that bars 4 and 7 will never take on any load. These are truss bars (hinged on both ends) and therefore only take axial load. For them to take an axial load, they would have to deform along their axes, but since both extremities are constrained, they can't deform and therefore will never take a load. So we can remove them from our model.
Also, the region where the load is applied can be trivially solved. The load is vertical and only bar 2 has a vertical component, so it will have to absorb the entirety of the external force. So the vertical component of the axial force in bar 2 is equal to -10 kN. The horizontal component will have to be such that the resultant force is parallel to the bar, so we can find that
$$\begin{gather}
\frac{-10}{h} = \tan\theta_2 = \frac{3}{-2} \\
\therefore h = 6.67\ \text{kN} \\
N_2 = \sqrt{10^2 + 6.67^2} = 12.02\ \text{kN (compression)}
\end{gather}$$
Now, if bar 2 is generating a horizontal force of 6.67 kN on that node, bar 1 must balance that out, generating a horizontal force of -6.67 kN, which puts the bar in tension.
We can now remove bars 1 and 2 from the model as well, replacing them with their internal forces. We therefore end up with the following model (the supports are hinged, so no need for the hinge "ball"):
Now, the horizontal load applied by bar 1 is simply absorbed by the support and we can mostly forget about it.
Since bars 3 and 6 have the same tangent (in modulus), the vertical force of 10 kN carried by bar 2 will be equally divided between them. Just as with bar 2, this will generate horizontal forces within them. Using the same method as above, we find that $h = \pm1.67\ \text{kN}$ (positive for bar 6, negative for bar 3). The resultant axial force in the bars due to this load is therefore equal to 5.27 kN (tension for bar 3, compression for bar 6). Since the horizontal components generated by bars 3 and 6 cancel themselves out, bar 5 is unaffected.
And now we get to the point where we get stuck: the 6.67 kN horizontal force applied by bar 2. Bars 3, 5 and 6 all have horizontal components and can therefore participate in absorbing this force. We therefore have to figure out how the force is parcelled out between them.
Bars can be replaced with springs which resist the displacement of the nodes. The stiffness of the springs for truss bars is equal to $K = \dfrac{EA}{L}$. The springs work according to Hooke's Law, which basically states that
$$F = K\delta$$
where $\delta$ is the deformation of the bar.
Now, the node will clearly only deform in the horizontal direction. But bars 3 and 6 are inclined, so their stiffness is only partial, so we need to get the horizontal component of that stiffness, which is equal to $\overline{K}_3 = \overline{K}_6 = \dfrac{EA}{L}\cos\theta$ (where $\theta$ is the angle of bars 3 and 6 with the horizontal axis. The sign doesn't matter for cosine). The deformation felt by bars 3 and 6 is also inclined, so we also have to get just the horizontal component of that deformation, which gives us $\overline{\delta}_3 = \overline{\delta}_6 = \delta\cos\theta$.
Therefore, Hooke's Law for bars 3 and 6 for horizontal deformations actually looks like:
$$F = K\delta\cos^2\theta = \frac{EA}{L}\cos^2\theta\delta$$
The total stiffness felt by the node is therefore equal to (assuming equal $EA$ for all bars):
$$\begin{align}
K &= K_3 + K_5 + K_6 \\
K_3 = K_6 &= \frac{EA}{L_3}\cos^2\theta \\
K_5 &= \frac{EA}{L_5}
\end{align}$$
So, what fraction of the horizontal force will go to bar 5?
$$\begin{align}
f_5 &= \dfrac{\dfrac{EA}{L_5}}{\dfrac{EA}{L_5} + 2\dfrac{EA}{L_3}\cos^2\theta} \\
f_5 &= \dfrac{\dfrac{1}{L_5}}{\dfrac{1}{L_5} + 2\dfrac{1}{L_3}\cos^2\theta} \\
f_5 &= \dfrac{1}{1 + 2\dfrac{1}{\sqrt{10}}\left(\dfrac{1}{\sqrt{10}}\right)^2} = 0.9405 \\
\therefore f_3 = f_6 &= \frac{1 - 0.9405}{2} = 0.0297
\end{align}$$
So, bar 5 gets 94.05% of the horizontal load and bars 3 and 6 get 2.97% each. Bar 5 therefore has an axial compression of $0.9405\times6.67 = 6.27\ \text{kN}$, while the horizontal component of the axial force in bars 3 and 6 is equal to $0.0297\times6.67 = 0.20\ \text{kN}$. Using the same method as used previously for bars 2, 3 and 6, we can also find the vertical components for bars 3 and 6 as equal to $v = 0.20\times\dfrac{\pm3}{1} = \pm0.60\ \text{kN}$ (positive for bar 3, negative for bar 6). This generates a resultant axial force on bars 3 and 6 of 0.63 kN (compression in both).
Adding the axial forces in bars 3 and 6 due to both vertical and horizontal components of the forces applied by bar 2, we get:
$$\begin{alignat}{2}
N_3 &= 5.27 - 0.63 = 4.64\ \text{kN}&&\text{ (tension)} \\
N_6 &= -5.27 - 0.63 = -5.90\ \text{kN}&&\text{ (compression)} \\
\end{alignat}$$
To summarize, the forces in each of the bars is:
$$\begin{alignat}{2}
N_1 &= 6.67\ \text{kN}&&\text{ (tension)} \\
N_2 &= 12.02\ \text{kN}&&\text{ (compression)} \\
N_3 &= 4.64\ \text{kN}&&\text{ (tension)} \\
N_4 &= 0.00\ \text{kN}&& \\
N_5 &= 6.27\ \text{kN}&&\text{ (compression)} \\
N_6 &= 5.90\ \text{kN}&&\text{ (compression)} \\
N_7 &= 0.00\ \text{kN}&& \\
\end{alignat}$$
For the reactions, we can just use these internal components.
At the top support, we have bar 1 and bar 3.
$$\begin{alignat}{2}
V_{top} &= 5.00 - 0.60 &&= 4.40\ \text{kN} \\
H_{top} &= 6.67+1.67-0.20 &&= 8.14\ \text{kN} \\
\end{alignat}$$
At the middle support, we only have bar 5, so it only has a horizontal component of -6.27 kN.
$$\begin{align}
V_{mid} &= 0.00\ \text{kN} \\
H_{mid} &= -6.27\ \text{kN} \\
\end{align}$$
At the bottom support, we only have bar 6.
$$\begin{alignat}{2}
V_{bot} &= 5.00 + 0.60 &&= 5.60\ \text{kN} \\
H_{bot} &= -1.67-0.20 &&= -1.87\ \text{kN} \\
\end{alignat}$$
You'll notice that these reactions satisfy the global equilibrium equations. And here's the computer model to check our work (allowing for errors of rounding):
Now, there is a way to trivially solve this problem by adopting a simplifying assumption. All you have to do is assume that bar 5 is axially rigid, meaning it will not deform at all. The validity of such an assumption is questionable if the bar all have the same $EA$. However it is less than a third of the length of bars 3 and 6, meaning it is more than three times as stiff. If that's enough for you to consider it rigid (I personally believe that's only acceptable for things at least one order of magnitude stiffer than its neighbors, but I don't know of any code "suggestions" on the matter), the solution becomes trivial.
Since bar 5 is rigid, the central node won't suffer any horizontal displacements from the horizontal force applied by bar 2. Therefore, bars 3 and 6 won't deform and won't be affected by that component, meaning bar 5 will absorb the entirety of that horizontal component. And that's that. Obviously, this will change the results for bars 3 and 6, as well as the reactions. For comparison, here's the result in this case:
All figures obtained via Ftool, a free 2D frame analysis tool
