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Câu hỏi: Mặt phẳng $\left( P \right)$ song song với mặt phẳng $\left( Q \right):x+2y+z=0$ và cách $D\left( 1;0;3 \right)$ một khoảng bằng $\sqrt{6}$ thì $\left( P \right)$ có phương trình là:
A. $\left[ \begin{aligned}
& x+2y+z+2=0 \\
& x+2y+z-2=0 \\
\end{aligned} \right.$
B. $\left[ \begin{aligned}
& x+2y-z-10=0 \\
& x+2y+z-2=0 \\
\end{aligned} \right.$
C. $\left[ \begin{aligned}
& x+2y+z+2=0 \\
& -x-2y-z-10=0 \\
\end{aligned} \right.$
D. $\left[ \begin{aligned}
& x+2y+z+2=0 \\
& x+2y+z-10=0 \\
\end{aligned} \right.$
A. $\left[ \begin{aligned}
& x+2y+z+2=0 \\
& x+2y+z-2=0 \\
\end{aligned} \right.$
B. $\left[ \begin{aligned}
& x+2y-z-10=0 \\
& x+2y+z-2=0 \\
\end{aligned} \right.$
C. $\left[ \begin{aligned}
& x+2y+z+2=0 \\
& -x-2y-z-10=0 \\
\end{aligned} \right.$
D. $\left[ \begin{aligned}
& x+2y+z+2=0 \\
& x+2y+z-10=0 \\
\end{aligned} \right.$
Do $\left( P \right)\text{//}\left( Q \right)\Rightarrow \left( P \right):x+2y+z+m=0$
Lại có: $d\left( D,\left( P \right) \right)=\sqrt{6}\Leftrightarrow \dfrac{\left| 1+2.0+3+m \right|}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{1}^{2}}}}=\sqrt{6}$
$\Rightarrow \dfrac{\left| m+4 \right|}{\sqrt{6}}=\sqrt{6}\Leftrightarrow \left| m+4 \right|=6\Leftrightarrow \left[ \begin{aligned}
& m=2 \\
& m=-10 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& \left( P \right):x+2y+z+2=0 \\
& \left( P \right):x+2y+z-10=0 \\
\end{aligned} \right.$
Lại có: $d\left( D,\left( P \right) \right)=\sqrt{6}\Leftrightarrow \dfrac{\left| 1+2.0+3+m \right|}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{1}^{2}}}}=\sqrt{6}$
$\Rightarrow \dfrac{\left| m+4 \right|}{\sqrt{6}}=\sqrt{6}\Leftrightarrow \left| m+4 \right|=6\Leftrightarrow \left[ \begin{aligned}
& m=2 \\
& m=-10 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& \left( P \right):x+2y+z+2=0 \\
& \left( P \right):x+2y+z-10=0 \\
\end{aligned} \right.$
Đáp án D.