Trong không gian với hệ trục tọa độ Oxyz cho Aleft( 1;2; - 1 right),,,Bleft( -

Gọi $M\left( {a;b;c} \right)$, $M \in \left( P \right) \Rightarrow a - 2b + c + 4 = 0$
$\overrightarrow {MA} = \left( {1 - a;2 - b; - 1 - c} \right),\overrightarrow {MB} = \left( { - 2 - a;1 - b; - c} \right)$
$\begin{array}{l}
MA = MB = 4 \Leftrightarrow \left\{ \begin{array}{l}
M{A^2} = M{B^2}\\
M{A^2} = \frac{{11}}{4}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {1 - a} \right)^2} + {\left( {2 - b} \right)^2} + {\left( { - 1 - c} \right)^2} = {\left( { - 2 - a} \right)^2} + {\left( {1 - b} \right)^2} + {\left( { - c} \right)^2}\\
{\left( {1 - a} \right)^2} + {\left( {2 - b} \right)^2} + {\left( { - 1 - c} \right)^2} = \frac{{11}}{4}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
1 - 2a + 4 - 4b + 1 + 2c = 4 + 4a + 1 - 2b\\
{\left( {1 - a} \right)^2} + {\left( {2 - b} \right)^2} + {\left( { - 1 - c} \right)^2} = \frac{{11}}{4}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
- 6a - 2b + 2c + 1 = 0\\
{\left( {1 - a} \right)^2} + {\left( {2 - b} \right)^2} + {\left( { - 1 - c} \right)^2} = \frac{{11}}{4}\,\,\,\left( * \right)
\end{array} \right.
\end{array}$
Từ hệ phương trình $\left\{ \begin{array}{l}
a - 2b + c + 4 = 0\\
- 6a - 2b + 2c + 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
7a - c + 3 = 0\\
2b = a + c + 4
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
c = 7a + 3\\
b = 4a + \frac{7}{2}
\end{array} \right.$
Thay vào phương trình $\left( * \right)$ ta được
$\begin{array}{l}
{\left( {1 - a} \right)^2} + {\left( { - 4a - \frac{3}{2}} \right)^2} + {\left( {7a + 4} \right)^2} = \frac{{11}}{4}\\
\Leftrightarrow 1 - 2a + {a^2} + 16{a^2} + 12a + \frac{9}{4} + 49{a^2} + 56a + 16 - \frac{{11}}{4} = 0\\
\Leftrightarrow 66{a^2} + 66a + \frac{{33}}{2} = 0 \Leftrightarrow a = - \frac{1}{2}.
\end{array}$