Charge on capacitors
If the three capacitors are connected in series so that they form a ring, then kirchhoff's voltage law must be satisfied such that the voltage across the three components as you traverse the loop must add up to 0.

V1 + V2 + V3 = 0


   - V1 +  - V2 +  - V3 + 
     C1      C2      C3
+----||------||------||---+ 
|                         |
+-------------------------+


If the capacitors initially had voltage V1, V2, V3 before they were connected then some charge Q would flow such that the final volages V1' + V2' + V3' = 0.

V1' + V2' + V3' = 0
V1' = V1 + Q/C1
V2' = V2 + Q/C2
V3' = V3 + Q/C3

Substituting the equations gives...
(V1 + Q/C1) + (V2 + Q/C2) + (V3 + Q/C3) = 0

Solving for Q gives...
Q = -(V1 + V2 + V3)/(1/C1 + 1/C2 + 1/C3)

V1' = V1 -(V1 + V2 + V3)/(1/C1 + 1/C2 + 1/C3)/C1
V2' = V2 -(V1 + V2 + V3)/(1/C1 + 1/C2 + 1/C3)/C2
V3' = V3 -(V1 + V2 + V3)/(1/C1 + 1/C2 + 1/C3)/C3


For example if C1=1F, C2=1F, C3=2F, V1=5V, V2=5V, V3=0V then

Q = -(5V + 5V + 0V) / (1/(1F) + 1/(1F) + 1/(2F)) = -4C.
V1' = 5V + -4C/1F = 1V
V2' = 5V + -4C/1F = 1V
V3' = 0V + -4C/2F = -2V

You could analyze the circuit as two series capacitors in parallel with the third, but the choice of which capacitors are in series or parallel is arbitrary.

The capacitors would stay charged indefinitely except for leakage current across the dielectric and insulation.