Đáp án:
$\begin{array}{l}
a)A\left( {1; - 2} \right);B\left( {2;4} \right);C\left( { - 3; - 1} \right)\\
\overrightarrow {AB} = \left( {1;6} \right);\overrightarrow {AC} = \left( { - 4;1} \right)\\
\Rightarrow \overrightarrow {AB} .\overrightarrow {AC} = 1.\left( { - 4} \right) + 6.1 = - 4 + 6 = 2\\
Vậy\,\overrightarrow {AB} .\overrightarrow {AC} = 2\\
b)D \in Ox\\
\Rightarrow D\left( {x;0} \right)\\
DA = DB\\
\Rightarrow D{A^2} = D{B^2}\\
\Rightarrow {\left( {1 - x} \right)^2} + {\left( { - 2} \right)^2} = {\left( {2 - x} \right)^2} + {4^2}\\
\Rightarrow {x^2} - 2x + 1 + 4 = {x^2} - 4x + 4 + 16\\
\Rightarrow 2x = 15\\
\Rightarrow x = \dfrac{{15}}{2}\\
\Rightarrow D\left( {\dfrac{{15}}{2};0} \right)
\end{array}$
