\[\begin{array}{l} {\text{а)}} \hfill \\ \frac{{AM}}{{AK}} = 3:\frac{7}{4} = \frac{{12}}{7}. \hfill \\ AD = \sqrt {A{B^2} - B{D^2}} = \sqrt {{{12}^2} - {6^2}} = 6\sqrt 3 \hfill \\ AO = \frac{2}{3}AD = 4\sqrt 3 \hfill \\ AS = \sqrt {A{O^2} + S{O^2}} = 7. \hfill \\ \frac{{AB}}{{AS}} = \frac{{12}}{7} = \frac{{AM}}{{AK}} \Rightarrow \vartriangle AKM \sim \vartriangle ASB \Rightarrow KM\parallel SB. \hfill \\ {\text{Аналогично}}{\text{, }}KN\parallel SC. \hfill \\ \left. \begin{array}{l} KM\parallel SB \hfill \\ KN\parallel SC \hfill \\ \end{array} \right\} \Rightarrow \left( {MNK} \right)\parallel \left( {SBC} \right). \hfill \\ {\text{б)}} \hfill \\ AP = \frac{1}{4}AD = \frac{{3\sqrt 3 }}{2} \hfill \\ PD = AD - AP = \frac{{9\sqrt 3 }}{2} \hfill \\ {S_{SPD}} = \frac{1}{2} \cdot \frac{{9\sqrt 3 }}{2} \cdot 1 = \frac{{9\sqrt 3 }}{4} \hfill \\ MN\parallel \left( {SBC} \right) \Rightarrow \rho \left( {M,SBC} \right) = \rho \left( {P,SBC} \right) = PQ \hfill \\ {S_{SPD}} = \frac{1}{2} \cdot PQ \cdot \sqrt {13} = \frac{{9\sqrt 3 }}{4} \Leftrightarrow PQ = \frac{{9\sqrt 3 }}{{2\sqrt {13} }} = \frac{{9\sqrt {39} }}{{26}}. \hfill \\ \end{array}\]

б) \[\frac{{9\sqrt {39} }}{{26}}\]