Đáp án:

$\sqrt{85}$

Giải thích các bước giải:

$A(1;-3;2); B(-2;1;-4)\\ z_Az_B<0$

$\Rightarrow A,B$ nằm về 2 phía mặt phẳng $(xOy)$

Dựng hình bình hành $AMNC, $

$\Rightarrow AC//MN; AC=MN; AM=CN\\ MN \subset (Oxy)\\ \Rightarrow AC //(Oxy) \Rightarrow z_C=z_A=2$

$MN=4, A$ cố định

$\Rightarrow C$ chạy trên đường tròn tâm $A$ bán kính $4cm$

$\Rightarrow$ Toạ độ $x_C, y_C$ thoả: $(x_C-1)^2+(y_C+3)^2=16(1)$

Lấy $B'$ đối xứng với $B$ qua mặt phẳng $(xOy)$

$\Rightarrow B'=(-2;1;4)\\ N \in (xOy) \Rightarrow BN=B'N\\ |AM-BN|=|CN-BN|=|CN-B'N| \le CB'$

Dấu "=" xảy ra $\Leftrightarrow B',C, N$ thẳng hàng.

$CB'^2=(x_C+2)^2+(y_C-1)^2+(4-2)^2\\ CB'^2-4=(x_C+2)^2+(y_C-1)^2\\ \sqrt{CB'^2-4}=\sqrt{(x_C+2)^2+(y_C-1)^2}\\ =\sqrt{(x_C-1+3)^2+(y_C+3-4)^2}\\ \le \sqrt{(x_C-1)^2+(y_C+3)^2}+\sqrt{3^3+4^2}(BĐT Mincopxki)\\ =9\\ \Rightarrow CB'^2-4 \le 81\\ \Rightarrow CB'^2 \le 85\\ \Rightarrow CB' \le \sqrt{85}$

Dấu "=" xảy ra $\Leftrightarrow  \dfrac{x_C-1}{y_C+3}=\dfrac{3}{-4}$

$\Leftrightarrow  -4x_C+4=3y_C+9\\ \Leftrightarrow  y_C=\dfrac{-4x_C-5}{3}$

Thế vào $(1)$

$\Rightarrow  (x_C-1)^2+\left(\dfrac{-4x_C-5}{3}+3\right)^2=16\\ \Rightarrow  \dfrac{25x_C^2}{9}-\dfrac{50x_C}{9}+\dfrac{25x}{9}=16\\ \Rightarrow 25x_C^2-50x_C+25=144\\ \Rightarrow 25x_C^2-50x_C-119=0\\ \Rightarrow  \left[\begin{array}{l} x_C=\dfrac{17}{5}\\ x_C=\dfrac{-7}{5}\end{array} \right.\\ \Rightarrow  \left[\begin{array}{l} x_C=\dfrac{17}{5}, y_C=\dfrac{-31}{5}\\ x_C=\dfrac{-7}{5}, y_C=\dfrac{1}{5}\end{array} \right.\\ \Rightarrow \left[\begin{array}{l} C=\left(\dfrac{17}{5};\dfrac{-31}{5};2\right)\\C=\left(\dfrac{-7}{5}; \dfrac{1}{5};2\right)\end{array} \right.\\ \circledast C=\left(\dfrac{17}{5};\dfrac{-31}{5};2\right)\\ \overrightarrow{B'C}= \left(\dfrac{27}{5};-\dfrac{36}{5};-2\right)=\dfrac{1}{5}\left(27;-36;-10\right)\\ \Rightarrow B'C: \left\{\begin{array}{l} x=-2+27t\\ y=1-36t\\z=4-10t\end{array} \right.\\ N=B'C \cap (xOy)=\left(\dfrac{44}{5};\dfrac{-67}{5};0\right)$

Do $AMNC$ là hình bình hành

$\Rightarrow \left\{\begin{array}{l} x_M=x_A+x_N-x_C=\dfrac{32}{5}\\y_M=y_A+y_N-y_C =\dfrac{-51}{5}\\z_M=z_A+z_N-z_C=0\end{array} \right.\\ \Rightarrow M=\left(\dfrac{32}{5};-\dfrac{51}{5};0\right)\\ \circledast C=\left(\dfrac{-7}{5}; \dfrac{1}{5};2\right)\\ N=\left(-\dfrac{4}{5};-\dfrac{3}{5};0\right)\\ M=\left(\dfrac{8}{5};-\dfrac{19}{5};0\right)$