Đáp án:
B
Giải thích các bước giải:
\(\begin{array}{l}\overrightarrow {AB} = \left( {1 - a;b;0} \right),\overrightarrow {AC} = \left( {1 - a;0;c} \right),\overrightarrow {BC} = \left( {0; - b;c} \right)\\\overrightarrow {AH} = \left( {3 - a;2;1} \right),\overrightarrow {BH} = \left( {2;2 - b;1} \right),\left[ {\overrightarrow {AB} ,\overrightarrow {AC} } \right] = \left( {bc,c\left( {a - 1} \right),b\left( {a - 1} \right)} \right)\end{array}\)
Điểm H là trực tâm tam giác ABC
\( \Leftrightarrow \left\{ \begin{array}{l}\overrightarrow {AH} .\overrightarrow {BC} = 0\\\overrightarrow {BH} .\overrightarrow {AC} = 0\\\left[ {\overrightarrow {AB} ,\overrightarrow {AC} } \right].\overrightarrow {AH} = 0\end{array} \right.\) \( \Leftrightarrow \left\{ \begin{array}{l} - 2b + c = 0\\2\left( {1 - a} \right) + c = 0\\bc\left( {3 - a} \right) + 2c\left( {a - 1} \right) + b\left( {a - 1} \right) = 0\end{array} \right.\)
\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}c = 2b\\a = 1 + b\\2{b^2}\left( {3 - 1 - b} \right) + 2.2b.b + b.b = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}c = 2b\\a = 1 + b\\ - 2{b^3} + 4{b^2} + 4{b^2} + {b^2} = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}c = 2b\\a = 1 + b\\ - 2{b^3} + 9{b^2} = 0\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}c = 2b\\a = 1 + b\\\left[ \begin{array}{l}b = 0\\b = \dfrac{9}{2}\end{array} \right.\end{array} \right. \Rightarrow \left[ \begin{array}{l}a = 1,b = 0,c = 0\left( {loai} \right)\\a = \dfrac{{11}}{2},b = \dfrac{9}{2},c = 9\end{array} \right.\\ \Rightarrow a + b + c = 19\end{array}\)
