Cho \( \left( {{ \Delta _1}} \right): \, \, \dfrac{{x - 1}}{5} = \dfrac{{y - 2}}{2} = \dfrac{z}{1}, \, \, \left( {{ \Delta _2}} \right): \, \, \dfrac{{x - 3}}{1} = \dfrac{{y - 6}}{1} = \dfrac{{z - 2}}{2} \). \( \left( d \right) \) qua \(O \) và vuông góc với \( \left( {{ \Delta _1}} \right), \, \, \left( {{ \Delta _2}} \right) \) có phương trình:
A.\(\left\{ \begin{array}{l}x = t\\y = - 3t\\z = - t\end{array} \right.\)
B.\(\left\{ \begin{array}{l}x = 2t\\y = 3t\\z = t\end{array} \right.\)
C.\(\left\{ \begin{array}{l}x = t\\y = 3t\\z = t\end{array} \right.\)
D.\(\left\{ \begin{array}{l}x = t\\y = - 3t\\z = t\end{array} \right.\)
A.\(\left\{ \begin{array}{l}x = t\\y = - 3t\\z = - t\end{array} \right.\)
B.\(\left\{ \begin{array}{l}x = 2t\\y = 3t\\z = t\end{array} \right.\)
C.\(\left\{ \begin{array}{l}x = t\\y = 3t\\z = t\end{array} \right.\)
D.\(\left\{ \begin{array}{l}x = t\\y = - 3t\\z = t\end{array} \right.\)
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Hóa Học lớp 12
