Câu hỏi 30

Gọi \(A\left( {a;0;0} \right) \in Ox\), \(B\left( {0;b;0} \right) \in Oy\), \(C\left( {0;0;c} \right) \in Oz\).

Ta có: \(\overrightarrow {SA}  = \left( {a - 4; - 2; - 2} \right)\), \(\overrightarrow {SB}  = \left( { - 4;b - 2; - 2} \right)\), \(\overrightarrow {SC}  = \left( { - 4; - 2;c - 2} \right)\).

\(\left\{ \begin{array}{l}\overrightarrow {SA} .\overrightarrow {SB}  = 0\\\overrightarrow {SB} .\overrightarrow {SC}  = 0\\\overrightarrow {SA} .\overrightarrow {SC}  = 0\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l} - 4\left( {a - 4} \right) - 2\left( {b - 2} \right) + 4 = 0\\16 - 2\left( {b - 2} \right) - 2\left( {c - 2} \right) = 0\\ - 4\left( {a - 4} \right) + 4 - 2\left( {c - 2} \right) = 0\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l} - 4a - 2b + 24 = 0\\ - 2b - 2c + 24 = 0\\ - 4a - 2c + 24 = 0\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l}a = 3\\b = 6\\c = 6\end{array} \right.\).

\( \Rightarrow A\left( {3;0;0} \right);\,\,B\left( {0;6;0} \right);\,\,C\left( {0;0;6} \right)\)

\(\begin{array}{l} \Rightarrow SA = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - 2} \right)}^2} + {{\left( { - 2} \right)}^2}}  = 3\\\,\,\,\,\,\,SB = \sqrt {{{\left( { - 4} \right)}^2} + {4^2} + {{\left( { - 2} \right)}^2}}  = 6\\\,\,\,\,\,\,SC = \sqrt {{{\left( { - 4} \right)}^2} + {{\left( { - 2} \right)}^2} + {4^2}}  = 6\end{array}\)

Vậy \({V_{S.ABC}} = \dfrac{1}{6}SA.SB.SC = \dfrac{1}{6}.3.6.6 = 18\).

Chọn A.