$\displaystyle \begin{array}{{>{\displaystyle}l}} \mathrm{Ta\ có\ d_{1} :( x-1) -2( y+3) =0\ hay\ {\displaystyle x-2y-6=0}}\\ \mathrm{1.d( A,d_{2}) =\frac{|3.-3-4.1+1|}{\sqrt{3^{2} +( -4)^{2}}} =\frac{11}{5}}\\ \\ \mathrm{d( B,d_{1}) =\frac{|1.1-2.3-6|}{\sqrt{1^{2} +( -2)^{2}}} =\frac{11}{\sqrt{5}}}\\ \mathrm{2.AB\ đi\ qua\ A( -3;1) \ nhận\ \overrightarrow{AB} \ ( 4;2) \ là\ vtcp\ hay\ \vec{n}( 1;-2) \ là\ vtpt}\\ \mathrm{AB:( x+3) -2( y-1) =0\ hay\ x-2y+5=0}\\ \mathrm{a.\ d_{1} \cap d_{2} \Leftrightarrow \{^{{\displaystyle x-2y-6=0}}_{3x-4y+1=0} \ \Leftrightarrow \ x=-13\ và\ y=\frac{-19}{2}}\\ \mathrm{vậy\ d_{1} \cap d_{2} =\left( -13;\frac{-19}{2}\right)}\\ \mathrm{b.\ \ Xét\ d_{1} \cap AB\ có\ \frac{1}{1} =\frac{-2}{-2} \neq \frac{-6}{5}}\\ \mathrm{\Rightarrow d_{1} //AB}\\ \mathrm{vậy\ d_{1} \cap d_{2} =\phi }\\ \mathrm{c.\ AB\cap d_{2} \Leftrightarrow \{^{{\displaystyle x-2y+5=0}}_{3x-4y+1=0} \ \Leftrightarrow \ x=9\ và\ y=7}\\ \mathrm{vậy\ AB\cap d_{2} =( 9;7)}\\ \mathrm{3.\ Ta\ có\ cos\ \widehat{d_{1} ,d_{2}} =|\frac{\overrightarrow{n_{1}} .\overrightarrow{n_{2}}}{\overrightarrow{|n_{1}} ||\overrightarrow{n_{2} |}} |=\frac{|1.3+( -2) .( -4) |}{\sqrt{1^{2} +( -2)^{2}}\sqrt{3^{2} +( -4)^{2}}} =\frac{24\sqrt{5}}{25}}\\ \mathrm{4.}\\ \mathrm{a.\ \Delta \ đi\ qua\ A( -3;1) \ và\ song\ song\ với\ d_{1} \ nhận\ \overrightarrow{u_{1}}( 2;1) \ là\ vtcp}\\ \mathrm{\Delta :\{^{x=-3+2t}_{y=1+t}}\\ \mathrm{b.\ \Delta '\ đi\ qua\ B( 1;3) \ và\ song\ song\ với\ d_{2} \ nhận\ \overrightarrow{u_{2}}( 4;3) \ là\ vtcp}\\ \mathrm{\Delta ':\{^{x=1+4t}_{y=3+3t}}\\ \mathrm{5.\ }\\ \mathrm{a.\ \Delta \ đi\ qua\ A( -3;1) \ và\ song\ song\ với\ d_{2} \ nhận\ \overrightarrow{n_{2}}( 3;-4) \ là\ vtpt}\\ \mathrm{\Delta :3( x+3) -4( y-1) =0\ hay\ \ 3x-4y+13=0}\\ \mathrm{b.\ \Delta '\ đi\ qua\ B( 1;3) \ và\ song\ song\ với\ d_{1} \ nhận\ \overrightarrow{n_{1}}( 1;-2) \ là\ vtpt}\\ \mathrm{\Delta ':\ ( x-1) -2( y-3) =0\ hay\ x-2y+5=0}\\ \mathrm{6.\ M\in d_{1} \ \Rightarrow M( 1+2a;a-3)}\\ \mathrm{a.\ Để\ \ \Delta AMB\ vuông\ tại\ A\ \Leftrightarrow \overrightarrow{AB} .\overrightarrow{AM} =0}\\ \mathrm{\Leftrightarrow 4( 2a+1+3) +2( a-3-1) =0}\\ \mathrm{\Leftrightarrow a=\frac{-4}{5} \ \Rightarrow M\left(\frac{-3}{5} ;\frac{-19}{5}\right)}\\ \mathrm{b.\ AM=7\ \Leftrightarrow AM^{2} =49\Leftrightarrow ( 2a+4)^{2} +( a-4)^{2} =49}\\ \mathrm{\Leftrightarrow 5a^{2} +8a-17=0\ \Leftrightarrow a=\frac{-4\pm \sqrt{101}}{5} \ \Rightarrow M( ...;...)}\\ \mathrm{c.\ Ta\ có\ \overrightarrow{MA} -\overrightarrow{2MB} =( -3-1-2a-2( 1-1-2a) ;1-a+3-2( 3-a+3)}\\ \mathrm{=( -4+2a;-8+a)}\\ \mathrm{\overrightarrow{|MA} -\overrightarrow{2MB} |=6\Leftrightarrow \sqrt{( 2a-4)^{2} +( a-8)^{2}} =6\ \Leftrightarrow 5a^{2} -32a+80=36}\\ \mathrm{\Leftrightarrow a=\frac{22}{5} \ hoặc\ a=2\Rightarrow M\left(\frac{49}{5} ;\frac{7}{5}\right) \ hoặc\ M( 5;-1)}\\ \mathrm{d.\overrightarrow{MA} +\overrightarrow{2MB} =( -3-1-2a+2( 1-1-2a) ;1-a+3+2( 3-a+3)}\\ \mathrm{=( -4-6a;16-3a)}\\ \mathrm{|\overrightarrow{MA} +\overrightarrow{2MB} |\ đạt\ GTNN\ \Leftrightarrow ( -6a-4)^{2} +( -3a+16)^{2} \ đạt\ GTNN}\\ \mathrm{\Leftrightarrow 45a^{2} -48a+272\ đạt\ GTNN}\\ \mathrm{Ta\ có\ 45a^{2} -48a+272=45\left( a^{2} -2.\frac{1}{2} .\frac{16}{15} a+\frac{64}{225}\right) +\frac{1296}{5} \leqslant \frac{1296}{5}}\\ \mathrm{Dấu\ "="\ xảy\ ra\ \Leftrightarrow a=\frac{8}{15}}\\ \mathrm{Vậy\ M\left(\frac{31}{15} ;\frac{-37}{15}\right)}\\ \mathrm{e.\ Ta\ có\ MA+MO\ đạt\ GTNN\ \Leftrightarrow M,\ A,\ O\ thẳng\ hàng.}\\ \mathrm{AO\ đi\ qua\ O( 0;0) \ nhận\ \vec{n}( 3;1) \ là\ vtpt}\\ \mathrm{AO:3x+y=0}\\ \mathrm{M,\ A,\ O\ thẳng\ hàng\ \Rightarrow M\in AO\ hay\ 3( 2a+1) +( a-3) =0}\\ \mathrm{\Leftrightarrow a=0}\\ \mathrm{Vậy\ M( 1;-3)}\\ \mathrm{f.\ d( M,d_{2}) =\frac{|3.( 2a+1) -4.( a-3) +1|}{\sqrt{3^{2} +( -4)^{2}}} =3}\\ \mathrm{\Leftrightarrow |2a+16|=15}\\ \mathrm{\Leftrightarrow 2a-16=15\ hoặc\ 2a-16=-15}\\ \mathrm{\Leftrightarrow a=\frac{31}{2} \ hoặc\ a=\frac{1}{2}}\\ \mathrm{Vậy\ M\left( 32;\frac{25}{2}\right) \ hoặc\ M\left( 2;\frac{-5}{2}\right)} \end{array}$