C1 = 3600 pF
C2 = 7200 pF
C3 = 11000 pF
capacitance is minimum when all are connected in
series
1/Cmin = 1/C1 + 1/C2 + 1/C3
1/Cmin = 1/3600 + 1/7200 + 1/11000
Cmin = 1970.15 pF <<<<<-------answer
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