Đáp án đúng: C
Giải chi tiết:Ta có \(\left\{ \begin{array}{l}M\left( {1 + u; - 1 + u;3 - 2u} \right)\\N\left( {1 + 3t; - 4;4 + t} \right)\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\overrightarrow {AM} = \left( {u;u - 3;4 - 2u} \right)\\\overrightarrow {AN} = \left( {3t; - 6;t + 5} \right)\end{array} \right.\)Vì \(A,\,\,M,\,\,N\) thẳng hàng nên \(\overrightarrow {AM} = k\overrightarrow {AN} \,\).\(\begin{array}{l} \Rightarrow \left\{ \begin{array}{l}u = k.3t\\u - 3 = - 6k\\4 - 2u = k\left( {t + 5} \right)\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}u = 3 - 6k\\3 - 6k = 3kt\\4 - 6 + 12k = kt + 5k\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}u = 3 - 6k\\1 - 2k = kt\\4 - 6 + 12k = 1 - 2k + 5k\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}u = 3 - 6k\\1 - 2k = kt\\9k = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}k = \dfrac{1}{3}\\u = 1\\t = 1\end{array} \right.\end{array}\)\( \Rightarrow \left\{ \begin{array}{l}\overrightarrow {AM} \left( {1; - 2;2} \right)\\\overrightarrow {AN} \left( {3; - 6;6} \right)\end{array} \right.\).Vậy \(A{M^2} + A{N^2} = \left( {1 + 4 + 4} \right) + \left( {9 + 36 + 36} \right) = 90\).Chọn C
