$\begin{array}{*{20}{l}}
{\overrightarrow {AB} {\rm{ \;}} = \left( {2;3} \right),\overrightarrow {CD} {\rm{ \;}} = \left( { - 5; - 5} \right);\overrightarrow {AC} {\rm{ \;}} = \left( {3;0} \right)}\\
{\overrightarrow {BC} {\rm{ \;}} = \left( {1; - 3} \right),\overrightarrow {AD} {\rm{ \;}} = \left( { - 4; - 8} \right)}\\
{a)\overrightarrow {AB} .\overrightarrow {CD} {\rm{ \;}} = 2.\left( { - 5} \right) + 3.\left( { - 5} \right) = {\rm{ \;}} - 25}\\
{\overrightarrow {BC} .\overrightarrow {AD} {\rm{ \;}} = 1.\left( { - 4} \right) + \left( { - 3} \right).\left( { - 8} \right) = 20}\\
{b)\cos \widehat {BAC} = \frac{{\overrightarrow {AB} .\overrightarrow {AC} }}{{AB.AC}} = \frac{6}{{\sqrt {13} .3}} = \frac{2}{{\sqrt {13} }}}\\
{ \Rightarrow \widehat {BAC} \approx {{56}^0}}\\
{c)M \in Ox \Rightarrow M\left( {x;0} \right)}\\
{\overrightarrow {MA} {\rm{ \;}} = \left( {1 - x;2} \right),\overrightarrow {MB} {\rm{ \;}} = \left( {3 - x;5} \right)}\\
{ \Rightarrow \overrightarrow {MA} .\overrightarrow {MB} {\rm{ \;}} = 0}\\
{ \Rightarrow \left( {1 - x} \right)\left( {3 - x} \right) + 10 = 0}\\
{ \Leftrightarrow {x^2} - 4x + 13 = 0\left( {VN} \right)}
\end{array}$
