问题
若已知空间曲线 $\Gamma$ 的一般式方程为 $\left\{\begin{array}{l} F(x, y, z)=0, \\ G(x, y, z)=0 \end{array}\right.$, 则在曲线 $\Gamma$ 上的点 $(x_{0}, y_{0}, z_{0})$ 处，曲面 $F(x, y, z)$ $=$ $0$ 和 $G(x, y, z)$ $=$ $0$ 的两个法向量 $n_{1}$ 和 $n_{2}$ 分别为：$\boldsymbol{n}_{1}$ $=$ $\left(F_{x}^{\prime}\left(x_{0}, y_{0}, z_{0} \right), F_{y}^{\prime}\left(x_{0}, y_{0}, z_{0} \right), F_{z}^{\prime}\left(x_{0}, y_{0}, z_{0}\right)\right)$
$\boldsymbol{n}_{1}$ $=$ $\left(G_{x}^{\prime}\left(x_{0}, y_{0}, z_{0} \right), G_{y}^{\prime}\left(x_{0}, y_{0}, z_{0} \right), G_{z}^{\prime}\left(x_{0}, y_{0}, z_{0}\right)\right)$
该点处的切向量为：
$\boldsymbol{\tau}$ $=$ $\boldsymbol{n}_{1} \times \boldsymbol{n}_{2}$
若记切向量 $\boldsymbol{\tau}$ $=$ $(A, B, C)$,
则曲线 $\Gamma$ 在点 $(x_{0}, y_{0}, z_{0})$ 处的法平面方程是多少？
选项
[A]. $A$ $\left(x-x_{0}\right)$ $+$ $B$ $\left(y-y_{0}\right)$ $+$ $C$ $\left(z-z_{0}\right)$ $=$ $0$[B]. $\frac{x-x_{0}}{A}$ $=$ $\frac{y-y_{0}}{B}$ $=$ $\frac{z-z_{0}}{C}$
[C]. $A$ $\left(x-x_{0}\right)$ $-$ $B$ $\left(y-y_{0}\right)$ $-$ $C$ $\left(z-z_{0}\right)$ $=$ $0$
[D]. $A$ $\left(x+x_{0}\right)$ $+$ $B$ $\left(y+y_{0}\right)$ $+$ $C$ $\left(z+z_{0}\right)$ $=$ $0$
$A$ $\left(x-x_{0}\right)$ $+$ $B$ $\left(y-y_{0}\right)$ $+$ $C$ $\left(z-z_{0}\right)$ $=$ $0$