Chọn phương án đúng nhất: \(\left\{ \begin{array}{l} \left( \beta \right) \supset d\\ \left( \alpha \right) \cap \left( \beta \right) = a\\ a \cap d = I \end{array} \right. \Rig...

A. \(\left\{ \begin{array}{l}
\left( \beta  \right) \supset d\\
\left( \alpha  \right) \cap \left( \beta  \right) = a\\
a \cap d = I
\end{array} \right. \Rightarrow d \cap \left( \alpha  \right) = I\)

B. \(\left\{ \begin{array}{l}
S = \left( \alpha  \right) \cap \left( \beta  \right)\\
a \subset \left( \beta  \right)\\
a//\left( \alpha  \right)
\end{array} \right. \Rightarrow \left( \alpha  \right) \cap \left( \beta  \right) = d\left( {d\,qua\,S} \right)\)

C. \(\left\{ \begin{array}{l}
S = \left( \alpha  \right) \cap \left( \beta  \right)\\
a \subset \left( \alpha  \right),b \subset \left( \beta  \right)\\
a//b
\end{array} \right. \Rightarrow \left( \alpha  \right) \cap \left( \beta  \right) = d\left( {d\,//b} \right)\)

D. \(\left\{ \begin{array}{l}
a \not\subset \left( \alpha  \right)\\
d \cap a = I
\end{array} \right. \Rightarrow d \cap \left( \alpha  \right) = I\)