Trong không gian với hệ trục tọa độ Oxyz, gọi $\left( P \right)$...

Vì đi qua hai điểm A, B suy ra $\left\{ \begin{aligned}
& a+b+c+2=0 \\
& b+2c+2=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& c=a \\
& b=-2a-2 \\
\end{aligned} \right.$$\left( 1 \right)C\xrightarrow[{}]{{}}mp\left( P \right)d\left( C;\left( P \right) \right)=\dfrac{\left| 2a-b+c+2 \right|}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}=\dfrac{3\sqrt{2}}{2}\left( 2 \right)\left( 1 \right)\left( 2 \right)\left| 5a+4 \right|=\dfrac{3\sqrt{2}}{2}\sqrt{{{a}^{2}}+{{\left( 2a+2 \right)}^{2}}+{{a}^{2}}}\Leftrightarrow 2{{\left( 5a+4 \right)}^{2}}=9\left( 6{{a}^{2}}+8a+4 \right)\Leftrightarrow a=1a=c=1b=-2a-2=-4\xrightarrow[{}]{{}}abc=-4$.