\(\overrightarrow{AB}=\left(-1;-1;2\right)\); \(\overrightarrow{CD}=\left(2;2;-4\right)=-2\overrightarrow{AB}\)
\(\Rightarrow ABCD\) là hình thang có \(CD=2AB=2\sqrt{6}\)
\(\overrightarrow{AD}=\left(3;0;-3\right)\Rightarrow cos\widehat{BAD}=\dfrac{-9}{3\sqrt{2}.\sqrt{6}}=-\dfrac{\sqrt{3}}{2}\)
\(\Rightarrow\widehat{A}=150^0\Rightarrow\widehat{D}=30^0\) ; \(AD=3\sqrt{2}\)
\(S_{ABCD}=\dfrac{AB+CD}{2}.AD.sin\widehat{D}=\dfrac{9\sqrt{3}}{2}\)
\(\Rightarrow d\left(S;\left(ABCD\right)\right)=\dfrac{3V}{S_{ABCD}}=3\sqrt{3}\)
\(\left[\overrightarrow{AB};\overrightarrow{AD}\right]=3.\left(1;1;1\right)\)
\(H\left(0;1;5\right)\)
Đường thẳng d qua H và vuông góc (ABCD) có pt: \(\left\{{}\begin{matrix}x=t\\y=1+t\\z=5+t\end{matrix}\right.\)
\(\Rightarrow S\left(t;1+t;5+t\right)\)
Phương trình (ABCD):
\(1\left(x-1\right)+1\left(y-2\right)+1\left(z-3\right)=0\)
\(\Leftrightarrow x+y+z-6=0\)
\(d\left(S;\left(ABCD\right)\right)=3\sqrt{3}\Leftrightarrow\dfrac{\left|t+1+t+5+t-6\right|}{\sqrt{3}}=3\sqrt{3}\)
\(\Leftrightarrow\left|t\right|=3\Rightarrow\left[{}\begin{matrix}t=3\\t=-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}S\left(3;6;8\right)\\S\left(-3;-2;2\right)\end{matrix}\right.\)
