Capacitors in circuits, quick answer needed please

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Capacitors with capacitances of 2.00 μF, 4.00 μF, and 6.00 μF are connected in parallel to a 200-V source. When switches S1 and S2 are closed, the potential difference across each capacitor remains equal to the source voltage of 200 V. After disconnecting and closing switch S3, the capacitors are reconfigured, leading to a redistribution of charge. The final charge on the leftmost plate of each capacitor depends on their capacitance values and the new circuit configuration. The orientation of the charged capacitors will reflect a positive and negative voltage reference on each terminal in the new setup.
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Capacitors 1, 2 and 3, have capacitances equal to 2.00 μF,
4.00 μF, and 6.00 μF, respectively. The capacitors are connected in parallel, and
the parallel combination is connected across the terminals of a 200-V source. The
capacitors are then disconnected from both the voltage source and each other, and
are connected to three switches as shown in Figure 24-42. (a) What is the
potential difference across each capacitor when switches S1 and S2 are closed but
switch S3 remains open? (b) After switch S3 is closed, what is the final charge on
the leftmost plate of each capacitor? (c) Give the final potential difference across
each capacitor after switch S3 is closed.

For part b, is would the capacitors be connected in series? Also if they're in series would the sum of all 3 voltages be 0 b/c no voltage source is connected?
 
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You are going to need to include Figure 24-42.
 
crap sorry i attached it
 

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When the circuit is reconfigured, how are the charged capacitors oriented? I would expect to see a "+" and "-" voltage reference on each capacitor terminal in the new circuit.
 
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