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Câu hỏi: Trong không gian $Oxyz$, cho hai điểm $A\left( 0; 0; -3 \right)$, $B\left( 2; 0; -1 \right)$ và mặt phẳng $\left( P \right): 3x-8y+7z-1=0$. Gọi $C\left( a; b;c \right)$ với $a>0$ là điểm thuộc mặt phẳng $\left( P \right)$ sao cho tam giác $ABC$ đều. Tổng $a+b+c$ bằng
A. $-7$.
B. $-3$.
C. $3$.
D. $7$.
A. $-7$.
B. $-3$.
C. $3$.
D. $7$.
$C{{A}^{2}}=C{{B}^{2}}\Leftrightarrow {{a}^{2}}+{{b}^{2}}+{{\left( c+3 \right)}^{2}}={{\left( a-2 \right)}^{2}}+{{b}^{2}}+{{\left( c+1 \right)}^{2}}\Leftrightarrow 4a+4c=-4\Leftrightarrow c=-1-a$.
$C\in \left( P \right)\Leftrightarrow 3a-8b+7c-1=0\Leftrightarrow 3a-8b+7\left( -1-a \right)-1=0\Leftrightarrow -4a-8b=8\Leftrightarrow a=-2b-2$
$\Rightarrow c=-1-\left( -2b-2 \right)=2b+1$.
$C{{A}^{2}}=A{{B}^{2}}\Leftrightarrow {{a}^{2}}+{{b}^{2}}+{{\left( c+3 \right)}^{2}}=8\Leftrightarrow {{\left( -2b-2 \right)}^{2}}+{{b}^{2}}+{{\left( 2b+4 \right)}^{2}}=8$ $\Leftrightarrow 9{{b}^{2}}+24b+12=0$
$\Leftrightarrow \left[ \begin{aligned}
& b=-2\Rightarrow C\left( 2; -2; -3 \right) \\
& b=-\dfrac{2}{3}\Rightarrow C\left( -\dfrac{2}{3}; -\dfrac{2}{3}; -\dfrac{1}{3} \right) \left( L \right) \\
\end{aligned} \right.$.
Vậy $a+b+c=-3$.
$C\in \left( P \right)\Leftrightarrow 3a-8b+7c-1=0\Leftrightarrow 3a-8b+7\left( -1-a \right)-1=0\Leftrightarrow -4a-8b=8\Leftrightarrow a=-2b-2$
$\Rightarrow c=-1-\left( -2b-2 \right)=2b+1$.
$C{{A}^{2}}=A{{B}^{2}}\Leftrightarrow {{a}^{2}}+{{b}^{2}}+{{\left( c+3 \right)}^{2}}=8\Leftrightarrow {{\left( -2b-2 \right)}^{2}}+{{b}^{2}}+{{\left( 2b+4 \right)}^{2}}=8$ $\Leftrightarrow 9{{b}^{2}}+24b+12=0$
$\Leftrightarrow \left[ \begin{aligned}
& b=-2\Rightarrow C\left( 2; -2; -3 \right) \\
& b=-\dfrac{2}{3}\Rightarrow C\left( -\dfrac{2}{3}; -\dfrac{2}{3}; -\dfrac{1}{3} \right) \left( L \right) \\
\end{aligned} \right.$.
Vậy $a+b+c=-3$.
Đáp án B.