Cho \(A \left( {1;2;3} \right), \, \, \left( {{ \Delta _1}} \right): \, \, \dfrac{x}{2} = \dfrac{{y - 1}}{1} = \dfrac{{z + 1}}{{ - 1}}, \, \, \left( {{ \Delta _2}} \right): \, \dfrac{{x - 1}}{1} = \dfrac{{y + 1}}{{ - 2}} = \dfrac{{z - 2}}{1} \). Lập phương trình \( \left( d \right) \) qua \(A, \) \(d \bot \left( {{ \Delta _1}} \right) \) và \(d \) cắt \( \left( {{ \Delta _2}} \right) \).
A.\(\left\{ \begin{array}{l}x = 1 - 2t\\y = 2 + t\\z = 3 - 3t\end{array} \right.\)
B.\(\left\{ \begin{array}{l}x = 1 + t\\y = 2 + 5t\\z = 3 - t\end{array} \right.\)
C.\(\left\{ \begin{array}{l}x = 1 - 3t\\y = 2 + t\\z = 3 + 5t\end{array} \right.\)
D.\(\left\{ \begin{array}{l}x = 1 - t\\y = 2 + 2t\\z = 3 + t\end{array} \right.\)
A.\(\left\{ \begin{array}{l}x = 1 - 2t\\y = 2 + t\\z = 3 - 3t\end{array} \right.\)
B.\(\left\{ \begin{array}{l}x = 1 + t\\y = 2 + 5t\\z = 3 - t\end{array} \right.\)
C.\(\left\{ \begin{array}{l}x = 1 - 3t\\y = 2 + t\\z = 3 + 5t\end{array} \right.\)
D.\(\left\{ \begin{array}{l}x = 1 - t\\y = 2 + 2t\\z = 3 + t\end{array} \right.\)
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