Đáp án đúng: A
Giải chi tiết:Ta có: \(\overrightarrow {AB} \left( { - 1; - 1;2} \right);\,\,\overrightarrow {AC} = \left( {1; - 2;1} \right)\) \( \Rightarrow \left[ {\overrightarrow {AB} ;\overrightarrow {AC} } \right] = \left( {3;3;3} \right)\).
\( \Rightarrow {S_{\Delta ABC}} = \dfrac{1}{2}\left| {\left[ {\overrightarrow {AB} ;\overrightarrow {AC} } \right]} \right| = \dfrac{1}{2}.\sqrt {{3^2} + {3^2} + {3^2}} = \dfrac{{3\sqrt 3 }}{2}\).
Ta có: \(\overrightarrow {DC} = \left( { - 2; - 2;4} \right);\,\,\overrightarrow {AB} = \left( { - 1; - 1;2} \right) \Rightarrow \overrightarrow {DC} = 2\overrightarrow {AB} \).
\( \Rightarrow AB//CD\) \( \Rightarrow ABCD\) là hình thang.
Ta có:
\(\begin{array}{l}{S_{ABCD}} = \dfrac{1}{2}d\left( {C;AB} \right).\left( {AB + CD} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{2}d\left( {C;AB} \right).\left( {AB + 2AB} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{1}{2}d\left( {C;AB} \right).3AB\\\,\,\,\,\,\,\,\,\,\,\,\,\, = 3{S_{\Delta ABC}} = \dfrac{{9\sqrt 3 }}{2}\end{array}\)
Ta có: \({V_{ABCD}} = \dfrac{1}{3}SH.{S_{ABCD}}\) \( \Rightarrow SH = \dfrac{{3{V_{S.ABCD}}}}{{{S_{ABCD}}}} = \dfrac{{3.\dfrac{{27}}{2}}}{{\dfrac{{9\sqrt 3 }}{2}}} = 3\sqrt 3 \).
Lại có \(H\) là trung điểm của \(CD \Rightarrow H\left( {0;1;5} \right)\).
Gọi \(S\left( {a;b;c} \right)\) ta có \(\overrightarrow {SH} = \left( { - a;1 - b;5 - c} \right)\).
Vì \(SH \bot \left( {ABCD} \right) \Rightarrow \overrightarrow {SH} = k\left[ {\overrightarrow {AB} ;\overrightarrow {AC} } \right] = \left( {3k;3k;3k} \right)\).
Khi đó ta có: \(SH = \sqrt {9{k^2} + 9{k^2} + 9{k^2}} = 3\sqrt 3 \left| k \right| = 3\sqrt 3 \Leftrightarrow k = \pm 1\).
TH1: \(k = 1\) \( \Rightarrow \overrightarrow {SH} = \left( {3;3;3} \right) \Rightarrow \left\{ \begin{array}{l} - a = 3\\1 - b = 3\\5 - c = 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = - 3\\b = - 2\\c = 2\end{array} \right.\) \( \Rightarrow {S_1}\left( { - 3; - 2;2} \right)\).
TH2: \(k = - 1\) \( \Rightarrow \overrightarrow {SH} = \left( { - 3; - 3; - 3} \right) \Rightarrow \left\{ \begin{array}{l} - a = - 3\\1 - b = - 3\\5 - c = - 3\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}a = 3\\b = 4\\c = 8\end{array} \right.\) \( \Rightarrow {S_2}\left( {3;4;8} \right)\).
Vậy trung điểm \(I\) của \({S_1}{S_2}\) là: \(I\left( {0;1;5} \right)\).
Chọn A.
