Đáp án đúng: C
Giải chi tiết:Gọi \(A\left( 3+t;4+t;-8-4t \right)\in \left( AB \right)\), mà \(A\in \left( \alpha \right)\Rightarrow 3+t-8-4t-1=0\Leftrightarrow -3t-6=0\Leftrightarrow t=-2\)
\(\Rightarrow A\left( 1;2;0 \right)\)
Gọi \(B\left( 3+t';4+t';-8-4t' \right)\in \left( AB \right)\,\,\left( 3+t'>0\Leftrightarrow t'>-3 \right)\) ta có:
\(\begin{array}{l}\,\,A{B^2} = {\left( {3 + t' - 1} \right)^2} + {\left( {4 + t' - 2} \right)^2} + {\left( { - 8 - 4t'} \right)^2} = {\left( {t' + 2} \right)^2} + {\left( {t' + 2} \right)^2} + {\left( { - 4t' - 8} \right)^2}\\ \Leftrightarrow 18 = 18t{'^2} + 72t' + 72 \Leftrightarrow \left[ \begin{array}{l}t = - 1\,\,\left( {tm} \right)\\t = - 3\,\,\left( {ktm} \right)\end{array} \right. \Leftrightarrow B\left( {2;3; - 4} \right)\end{array}\)
Khi đó ta có:
\(\overrightarrow{AC}=\left( a-1;b-2;c \right),\overrightarrow{BC}=\left( a-2;b-3;c+4 \right)\)
\(\Delta ABC\) vuông tại C nên \(\overrightarrow{AC}.\overrightarrow{BC}=0\Leftrightarrow \left( a-1 \right)\left( a-2 \right)+\left( b-2 \right)\left( b-3 \right)+c\left( c+4 \right)=0\) \(\Leftrightarrow {{a}^{2}}-3a+2+{{b}^{2}}-5b+6+{{c}^{2}}+4c=0\,\,\,\left( 1 \right)\)
Vì \(\widehat{ABC}={{60}^{0}}\Rightarrow BC=AB.\cos {{60}^{0}}=\frac{3\sqrt{2}}{2}\) \(\Leftrightarrow B{{C}^{2}}={{\left( a-2 \right)}^{2}}+{{\left( b-3 \right)}^{2}}+{{\left( c+4 \right)}^{2}}=\frac{9}{2}\,\,\,\left( 2 \right)\)
Mà \(C\in \left( \alpha \right)\Rightarrow a+c-1=0\Leftrightarrow c=1-a\).
Thay c = 1 – a vào (1) ta có:
\(\begin{align} & \,\,\,\,\,{{a}^{2}}-3a+2+{{b}^{2}}-5b+6+{{\left( 1-a \right)}^{2}}+4\left( 1-a \right)=0 \\ & \Leftrightarrow {{a}^{2}}-3a+2+{{b}^{2}}-5b+6+{{a}^{2}}-2a+1+4-4a=0 \\ & \Leftrightarrow 2{{a}^{2}}+{{b}^{2}}-9a-5b+13=0\,\,\,\left( 3 \right) \\ \end{align}\)
Thay c = 1 – a vào (2) ta có:
\(\begin{align} & \,\,\,\,\,{{\left( a-2 \right)}^{2}}+{{\left( b-3 \right)}^{2}}+{{\left( c+4 \right)}^{2}}=\frac{9}{2} \\ & \Leftrightarrow {{a}^{2}}-4a+4+{{b}^{2}}-6b+9+{{a}^{2}}-10a+25=\frac{9}{2} \\ & \Leftrightarrow 2{{a}^{2}}+{{b}^{2}}-14a-6b+\frac{67}{2}=0\,\,\,\,\left( 4 \right) \\ \end{align}\)
\(\left( 3 \right)-\left( 4 \right)\Leftrightarrow 5a+b=\frac{41}{2}\Leftrightarrow b=\frac{41}{2}-5a\)
Thay ngược lại vào (3) ta có:
\(\begin{align} & 2{{a}^{2}}+{{\left( \frac{41}{2}-5a \right)}^{2}}-9a-5\left( \frac{41}{2}-5a \right)+13=0 \\ & \Leftrightarrow 27{{a}^{2}}-189a+\frac{1323}{4}=0\Leftrightarrow a=\frac{7}{2} \\ & \Rightarrow b=3,\,\,c=-\frac{5}{2} \\ & \Rightarrow a+b+c=\frac{7}{2}+3-\frac{5}{2}=4 \\ \end{align}\)
Chọn C.
