Chọn C$\sqrt{29}$

PTTS của BC là:

                    $d:\left\{ \begin{align}  & x=t \\  & y=3-t \\  & z=1-t \\ \end{align} \right.$

 Vì

                                                        $M\in d\Rightarrow M(t;3-t;1-t)$

$\begin{align}  & \overrightarrow{MC}=(-3-t;3+t;3+t) \\  & \overrightarrow{MB}=(-t;t;t) \\ \end{align}$

$MC=2MB$

$\Leftrightarrow \sqrt{{{(-3-t)}^{2}}+{{(3+t)}^{2}}+{{(3+t)}^{2}}}=2\sqrt{{{(-t)}^{2}}+{{t}^{2}}+{{t}^{2}}}$

$\Leftrightarrow 9{{t}^{2}}-18t-27=0$

$\Leftrightarrow \left[ \begin{matrix}   t=3  \\   t=-1  \\\end{matrix} \right.$

$\begin{align}  & t=3\Rightarrow M(3;0;2) \\  & \Rightarrow \overrightarrow{AM}=(1;0;2)\Rightarrow AM=\sqrt{5} \\ \end{align}$

$\begin{align}  & t=-1\Rightarrow M(-1;4;2) \\  & \Rightarrow \overrightarrow{AM}=(-3;4;2)\Rightarrow AM=\sqrt{29} \\ \end{align}$

$\Rightarrow AM=\sqrt{29}$