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，则空间直角坐标系下平面方程的三点式如何表示？选项
[A]. $\begin{vmatrix} x+x_{1} & y+y_{1} & z+z_{1} \\ x_{2}+x_{1} & y_{2}+y_{1} & z_{2}+z_{1} \\ x_{3}+x_{1} & y_{3}+y_{1} & z_{3}+z_{1} \end{vmatrix}$ $=$ $0$[B]. $\begin{vmatrix} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{vmatrix}$ $=$ $1$
[C]. $\begin{vmatrix} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{vmatrix}$ $=$ $0$
[D]. $\begin{vmatrix} x_{1}-x & y_{1}-y & z_{1}-z \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{2} & y_{3}-y_{2} & z_{3}-z_{2} \end{vmatrix}$ $=$ $0$
$\begin{vmatrix} \textcolor{red}{x}-x_{\textcolor{cyan}{1}} & \textcolor{red}{y}-y_{\textcolor{cyan}{1}} & \textcolor{red}{z}-z_{\textcolor{cyan}{1}} \\ x_{\textcolor{orange}{2}}-x_{\textcolor{cyan}{1}} & y_{\textcolor{orange}{2}}-y_{\textcolor{cyan}{1}} & z_{\textcolor{orange}{2}}-z_{\textcolor{cyan}{1}} \\ x_{\textcolor{yellow}{3}}-x_{\textcolor{cyan}{1}} & y_{\textcolor{yellow}{3}}-y_{\textcolor{cyan}{1}} & z_{\textcolor{yellow}{3}}-z_{\textcolor{cyan}{1}} \end{vmatrix}$ $=$ $\textcolor{red}{0}$