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Câu hỏi: Trong không gian $Oxyz$, cho bốn điểm $A\left( a;0;0 \right),B\left( 0;b;0 \right),C\left( 0;0;c \right),D\left( 1;2;-1 \right)$, với $a,b,c$ là các số thực khác $0$. Biết rằng bốn điểm $A,B,C,D$ đồng phẳng, khi khoảng cách từ gốc toạ độ $O$ đến mặt phẳng $\left( ABC \right)$ lớn nhất. Giá trị $a+b+c$ bằng
A. $15$.
B. $3$.
C. $2$.
D. $4$.
A. $15$.
B. $3$.
C. $2$.
D. $4$.
Phương trình mặt phẳng $\left( ABC \right):\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\text{ }\left( acb\ne 0 \right)$
Vì $D\left( 1;2;-1 \right)$ thuộc mặt phẳng nên ta có: $\dfrac{1}{a}+\dfrac{2}{b}-\dfrac{1}{c}=1$
Ta có $d\left( O;\left( ABC \right) \right)=\dfrac{1}{\sqrt{\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}}}$
Mà ${{\left( \dfrac{1}{a}+\dfrac{2}{b}-\dfrac{1}{c} \right)}^{2}}={{\left( \dfrac{1}{a}\text{+2}\text{.}\dfrac{1}{b}\text{+}\left( -1 \right)\text{.}\dfrac{1}{c} \right)}^{2}}\le 6\left( \dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}} \right)$
$\Rightarrow 1\le \sqrt{6}.\sqrt{\left( \dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}} \right)}\Rightarrow \dfrac{1}{\sqrt{\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}}}\le \sqrt{6}$ hay $d\left( O;\left( ABC \right) \right)\le \sqrt{6}$
Vậy, ${{d}_{\max }}=\sqrt{6}\Leftrightarrow \left\{ \begin{aligned}
& a=2b=-c \\
& \dfrac{1}{a}+\dfrac{2}{b}-\dfrac{1}{c}=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=6 \\
& b=3 \\
& c=-6 \\
\end{aligned} \right.\Rightarrow a+b+c=3$.
Vì $D\left( 1;2;-1 \right)$ thuộc mặt phẳng nên ta có: $\dfrac{1}{a}+\dfrac{2}{b}-\dfrac{1}{c}=1$
Ta có $d\left( O;\left( ABC \right) \right)=\dfrac{1}{\sqrt{\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}}}$
Mà ${{\left( \dfrac{1}{a}+\dfrac{2}{b}-\dfrac{1}{c} \right)}^{2}}={{\left( \dfrac{1}{a}\text{+2}\text{.}\dfrac{1}{b}\text{+}\left( -1 \right)\text{.}\dfrac{1}{c} \right)}^{2}}\le 6\left( \dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}} \right)$
$\Rightarrow 1\le \sqrt{6}.\sqrt{\left( \dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}} \right)}\Rightarrow \dfrac{1}{\sqrt{\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}}}\le \sqrt{6}$ hay $d\left( O;\left( ABC \right) \right)\le \sqrt{6}$
Vậy, ${{d}_{\max }}=\sqrt{6}\Leftrightarrow \left\{ \begin{aligned}
& a=2b=-c \\
& \dfrac{1}{a}+\dfrac{2}{b}-\dfrac{1}{c}=1 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=6 \\
& b=3 \\
& c=-6 \\
\end{aligned} \right.\Rightarrow a+b+c=3$.
Đáp án B.