I'll assume you know how to find the reactions but you're just unsure of the two hinges at C and E as that seems like your main concern. If you're not sure how to calculate the reactions I can add this later. I have used SkyCiv Beam to find the reactions:
As you can see these reactions balance just fine: $$\sum F_y = 11 + 10 + 5 - (6+2+6+2\times6) = 26 - 26 = 0\text{ kN} \ \sum M_A = -32 +6(2)+2(4)+6(5)+12(11)- 10(8) -5(14)= 0\text{ kN.m}$$
Now it doesn't really matter whether you choose to include the 2 kN point load at hinge C on member AC or CE. Just include it in the free body diagram (FBD) for one member or the other (NOT both!).
Let's make the 2 kN point load at C act on the right end of member AC, not the left end of member CE. Remembering that a moment can NOT be supported at the hinge C:
$$\sum F_y = 0\ 11 - 6 - 2 + H_C = 0\ \therefore H_C = 3\text{ kN}$$
Now consider member CE (again no moment at C or E). The force Hc needs to be in the opposite direction as that found in the FBD for member AC:
$$\sum F_y = 0\ H_C + H_E -6 = 0\ 3 + H_E - 6 = 0\ \therefore H_E = 3\text{ kN}$$
Lastly consider member <PERSON> to confirm that it all balances fine (again the force at E needs to be opposite that in the FBD for member CE):
$$\sum F_y = -H_E + 10 + 5 - 12 = -3 + 10 + 5 - 12 = 0 \text{ }\checkmark$$
Let's look at the shear force diagram (SFD) below and understand why it doesn't really matter which member the 2 kN point load acts on. 7
We solved earlier that at point C the shear force was Hc = 3 kN. As you can see in the SFD there are TWO values at point C (x=4m): 5 kN and 3 kN. Obviously the difference between these values is the 2 kN point load. If we had added the point load in our diagram for member CE instead of member AC then we would have solved the shear force at point C to be Hc = 5 kN. So you can include it in either member and it will be correct - just don't include it in both members.
SkyCiv Beam is pretty handy for analyses like this and it's a good way to check your logic, answers and working out. It will also solve the bending moment diagram (BMD) if you need it plus deflection, stress among others. 7