已知\(\left\{ {{a_n}} \right\}\)是等差数列，其前\(n\)项和为\({S_n}\)，若\({a_3} + 12{a_7} = {S_{12}}\)，则下列判断正确的是（       ）

因为\(\left\{ {{a_n}} \right\}\)是等差数列，\({a_3} + 12{a_7} = {S_{12}}\)，所以\({a_1} + 2d + 12\left( {{a_1} + 6d} \right) = 12{a_1} + 66d\)，

即\({a_1} + 8d = 0\)，亦即\({a_9} = 0\)，

所以\({S_{17}} = \frac{{\left( {{a_1} + {a_{17}}} \right) \times 17}}{2} = 0,{S_{11}} - {S_6} = {a_7} + {a_8} + {a_9} + {a_{10}} + {a_{11}} = 5{a_9} = 0\)，\({a_8} \cdot {a_{10}} = \left( {{a_9} - d} \right) \cdot \left( {{a_9} + d} \right) = a_9^2 - {d^2} = - {d^2} \leqslant 0\)．

故选：ABC．

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