问题
若平面过 $(x_{1}, y_{1}, z_{1})$, $(x_{2}, y_{2}, z_{2})$ 和 $(x_{3}, y_{3}, z_{3})$ 这三个点，则在空间直角坐标系下平面的 [法向量] $\vec{n}$ $=$ $?$选项
[A]. $\vec{n}$ $=$ $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ x_{1}-x_{2} & y_{1}-y_{2} & z_{1}-z_{2} \\ x_{1}-x_{3} & y_{1}-y_{3} & z_{1}-z_{3} \end{vmatrix}$[B]. $\vec{n}$ $=$ $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ x_{2}+x_{1} & y_{2}+y_{1} & z_{2}+z_{1} \\ x_{3}+x_{1} & y_{3}+y_{1} & z_{3}+z_{1} \end{vmatrix}$
[C]. $\vec{n}$ $=$ $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ x_{2}-x_{3} & y_{2}-y_{3} & z_{2}-z_{3} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{vmatrix}$
[D]. $\vec{n}$ $=$ $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{vmatrix}$
$\textcolor{red}{\vec{n}}$ $=$ $\begin{vmatrix} \textcolor{red}{\mathbf{i}} & \textcolor{red}{\mathbf{j}} & \textcolor{red}{\mathbf{k}} \\ x_{\textcolor{orange}{2}}-x_{\textcolor{cyan}{1}} & y_{\textcolor{orange}{2}}-y_{\textcolor{cyan}{1}} & z_{\textcolor{orange}{2}}-z_{\textcolor{cyan}{1}} \\ x_{\textcolor{yellow}{3}}-x_{\textcolor{cyan}{1}} & y_{\textcolor{yellow}{3}}-y_{\textcolor{cyan}{1}} & z_{\textcolor{yellow}{3}}-z_{\textcolor{cyan}{1}} \end{vmatrix}$