E là trung điểm AD', ta có \(E(0;\frac{7}{2};\frac{5}{2})\)
F là trung điểm B'C, ta có \(F(1;\frac{3}{2};\frac{1}{2})\)
\(\overrightarrow{EF}=(1;-2;-2)\)
\(\overrightarrow{AB}=\overrightarrow{EF}\)
\(\Leftrightarrow \left\{\begin{matrix} x_B+3=1\\ y_B-2=-2\\ z_B-1=-2 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x_B=-2\\ y_B=0\\ z_B=-1 \end{matrix}\right.\Rightarrow B(-2;0;-1)\)

\(\overrightarrow{A'B'}=\overrightarrow{EF}\)
\(\Leftrightarrow \left\{\begin{matrix} -2-x_{A'}=1\\ 1-y_{A'}=-2\\ 1-z_{A'}=-2 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x_{A'}=-3\\ y_{A'}=3\\ z_{A'}=3 \end{matrix}\right.\Rightarrow A'(-3;3;3)\)
\(\overrightarrow{D'C'}=\overrightarrow{EF}\)
\(\Leftrightarrow \left\{\begin{matrix} x_{C'}-3=1\\ y_{C'}-5=-2\\ z_{C'}-4=-2 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x_{C'}=4\\ y_{C'}=3\\ z_{C'}=2 \end{matrix}\right.\Rightarrow C'(4;3;2)\)
\(\overrightarrow{DC}=\overrightarrow{EF}\)
\(\left\{\begin{matrix} 4-x_D=1\\ 2-y_D=-2\\ 0-z_D=-2 \end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x_D=3\\ y_D=4\\ z_D=2 \end{matrix}\right.\Rightarrow D(3;4;2)\)
K là tâm hình hộp nên K là trung điểm AC'
Vậy \(K(\frac{1}{2};\frac{5}{2};\frac{3}{2})\)
b)
\(M(x;y;z)\in AA'\), ta có \(\overrightarrow{AA'}=(0;1;2)\)
\(\overrightarrow{AM}=k.\overrightarrow{AA'}\)
\(\left\{\begin{matrix} x+3=0\\ y-2=k\\ z-1=2k \end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=-3\\ y=2+k\\ z=1+2k \end{matrix}\right.\Rightarrow M(-3;2+k;1+2k)\)
\(\overrightarrow{KM}=(-\frac{7}{2};k-\frac{1}{2};2k-\frac{1}{2})\)
\(KM=\frac{\sqrt{59}}{2}\Leftrightarrow KM^2=\frac{59}{4}\)
\(\Leftrightarrow \frac{49}{4}+(k-\frac{1}{2})^2+(2k-\frac{1}{2})^2=\frac{59}{4}\)
\(\Leftrightarrow \frac{49}{4}+5k^2-3k+\frac{2}{4}=\frac{59}{4}\)
\(\Leftrightarrow 5k^2-3k-2=0\)
\(\Leftrightarrow \bigg \lbrack \begin{matrix} k=1\\ k=-\frac{2}{5} \end{matrix}\)
k = 1 → M(-3;3;3)
\(k=-\frac{2}{5}\rightarrow M(-3;\frac{8}{5};\frac{1}{5})\)