Trong không gian với hệ toạ độ $Oxyz$, cho hai đường thẳng...

Ta có :
${{\Delta }_{1}}: \dfrac{x}{1}=\dfrac{y-4}{2}=\dfrac{z-1}{3}\Leftrightarrow \left\{ \begin{aligned}
& x=a \\
& y=4+2a \\
& z=1+3a \\
\end{aligned} \right. \left( a\in \mathbb{R} \right).$${{\Delta }_{2}}: \dfrac{x+2}{-1}=\dfrac{y}{-2}=\dfrac{z-1}{3}\Rightarrow \left\{ \begin{aligned}
& x=-2-b \\
& y=-2b \\
& z=1+3b \\
\end{aligned} \right. \left( b\in \mathbb{R} \right) .MM\left\{ \begin{aligned}
& a=-2-b \\
& 4+2\text{a}=-2b \\
& 1+3\text{a}=1+3b \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-1 \\
& b=-1 \\
\end{aligned} \right. \Rightarrow M\left( -1 ;2 ; -2 \right).{{\Delta }_{1}}A\left( 1;6 ; 4 \right)\Rightarrow \overrightarrow{MA}=\left( 2 ; 4 ; 6 \right){{\Delta }_{2}}B\left( -2-b ;-2b ; 1+3b \right)MA=MB\Leftrightarrow M{{A}^{2}}=M{{B}^{2}}\Leftrightarrow 56={{\left( -1-b \right)}^{2}}+{{\left( -2b-2 \right)}^{2}}+{{\left( 3+3b \right)}^{2}}\Leftrightarrow 14{{b}^{2}}+28b-42=0\Leftrightarrow {{b}^{2}}+2b-3=0\Leftrightarrow \left[ \begin{aligned}
& b=1 \\
& b=-3 \\
\end{aligned} \right.\Rightarrow \left[ \begin{aligned}
& B\left( -3 ;-2 ; 4 \right) \\
& B\left( 1 ;6 ; -8 \right) \\
\end{aligned} \right.\Rightarrow \left[ \begin{aligned}
& \overrightarrow{MB}\left( -2 ;-4 ; 6 \right) \\
& \overrightarrow{MB}\left( 2 ;4 ; -6 \right) \\
\end{aligned} \right.\overrightarrow{MA}.\overrightarrow{MB}d\overrightarrow{MA}.\overrightarrow{MB} >0B\left( -3 ;-2 ; 4 \right)d\overrightarrow{u}=\overrightarrow{MA}+\overrightarrow{MB}=(0;0;12)\overrightarrow{u}dk\overrightarrow{u} \left( k\ne 0 \right)dk=\dfrac{-1}{12}d\overrightarrow{u}=\left( 0 ; 0 ; -1 \right)B$