Trong không gian $Oxyz,$ cho hai đường thẳng...

Cách giải:
Ta có $\left\{ \begin{aligned}
& M\left( 1+u;-1+u;3-2u \right) \\
& N\left( 1+3t;-4;4+t \right) \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& \overrightarrow{AM}=\left( u;u-3;4-2u \right) \\
& \overrightarrow{AN}=\left( 3t;-6;t+5 \right) \\
\end{aligned} \right.$
Vì $A,M,N$ thẳng hàng nên $\overrightarrow{AM}=k\overrightarrow{AN}.$
$\Rightarrow \left\{ \begin{aligned}
& u=k.3t \\
& u-3=-6k \\
& 4-2u=k\left( t+5 \right) \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& u=3-6k \\
& 3-6k=3kt \\
& 4-6+12k=kt+5k \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& u=3-6k \\
& 1-2k=kt \\
& 4-6+12k=1-2k+5k \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& u=3-6k \\
& 1-2k=kt \\
& 9k=3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& k=\dfrac{1}{3} \\
& u=1 \\
& t=1 \\
\end{aligned} \right.$
$\Rightarrow \left\{ \begin{aligned}
& \overrightarrow{AM}\left( 1;-2;2 \right) \\
& \overrightarrow{AN}\left( 3;-6;6 \right) \\
\end{aligned} \right..$
Vậy $A{{M}^{2}}+A{{N}^{2}}=\left( 1+4+4 \right)+\left( 9+36+36 \right)=90.$