问题
若两个平面的法向量分别为 $\vec{n_{1}}$ $=$ $(A_{1}, B_{1}, C_{1})$, $\vec{n_{2}}$ $=$ $(A_{2}, B_{2}, C_{2})$, 此外还有常数 $D_{1}$ 和 $D_{2}$, 则这两个平面相交所形成的直线如何表示？选项
[A]. $\left\{\begin{matrix} A_{1}x + B_{1}y + C_{1}z + D_{1} = 0,\\ A_{2}x + B_{2}y + C_{2}z + D_{2} = 0.\end{matrix}\right.$[B]. $\left\{\begin{matrix} \frac{A_{1}}{x} + \frac{B_{1}}{y} + \frac{C_{1}}{z} + D_{1} = 0,\\ \frac{A_{2}}{x} + \frac{B_{2}}{y} + \frac{C_{2}}{z} + D_{2} = 0.\end{matrix}\right.$
[C]. $\left\{\begin{matrix} \frac{1}{A_{1}}x + \frac{1}{B_{1}}y + \frac{1}{C_{1}}z + \frac{1}{D_{1}} = 0,\\ \frac{1}{A_{2}}x + \frac{1}{B_{2}}y + \frac{1}{C_{2}}z + \frac{1}{D_{2}} = 0.\end{matrix}\right.$
[D]. $\left\{\begin{matrix} A_{1}x + B_{1}y + C_{1}z = 0,\\ A_{2}x + B_{2}y + C_{2}z = 0.\end{matrix}\right.$
$\left\{\begin{matrix} A_{\textcolor{red}{1}}\textcolor{yellow}{x} + B_{\textcolor{red}{1}}\textcolor{yellow}{y} + C_{\textcolor{red}{1}}\textcolor{yellow}{z} + D_{\textcolor{red}{1}} = 0,\\ A_{\textcolor{cyan}{2}}\textcolor{yellow}{x} + B_{\textcolor{cyan}{2}}\textcolor{yellow}{y} + C_{\textcolor{cyan}{2}}\textcolor{yellow}{z} + D_{\textcolor{cyan}{2}} = 0.\end{matrix}\right.$