Đáp án:
a)
$\begin{array}{l}
\overrightarrow {AB} = \left( {10; - 2} \right)\\
\overrightarrow {AC} = \left( {4; - 3} \right)\\
Ta\,thấy\,:\frac{{10}}{4} = \frac{5}{2} \ne \frac{{ - 2}}{{ - 3}}\\
\Rightarrow \overrightarrow {AB} ;\overrightarrow {AC} \,ko\,cùng\,phương\\
\Rightarrow A,B,C\,ko\,thẳng\,hàng
\end{array}$
Vậy A,B,C tạo thành 1 tam giác
b)
$\begin{array}{l}
D\left( {x;y} \right) \Rightarrow \overrightarrow {AD} = \left( {x + 3;y - 3} \right)\\
\overrightarrow {AB} = \left( {10; - 2} \right)\\
\overrightarrow {AC} = \left( {4; - 3} \right)\\
Do:3\overrightarrow {AB} - 2\overrightarrow {AC} + 4\overrightarrow {AD} = \overrightarrow 0 \\
\Rightarrow \left\{ \begin{array}{l}
3.10 - 2.4 + 4\left( {x + 3} \right) = 0\\
3.\left( { - 2} \right) - 2.\left( { - 3} \right) + 4\left( {y - 3} \right) = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = - \frac{{37}}{5}\\
y = 3
\end{array} \right.\\
\Rightarrow D\left( { - \frac{{37}}{5};3} \right)
\end{array}$
