Giải thích các bước giải:
$\begin{array}{l}
\left\{ \begin{array}{l}
{\left( {x + 2} \right)^2} + {\left( {y - 5} \right)^2} + {\left( {z + 10} \right)^2} = 54\\
2x + 5y + z - 11 = 0\\
2x - y + z + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + 2} \right)^2} + {\left( {y - 5} \right)^2} + {\left( {z + 10} \right)^2} = 54\\
2x - y + z + 1 = 0\\
\left( {2x + 5y + z - 11} \right) - \left( {2x - y + z + 1} \right) - 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + 2} \right)^2} + {\left( {y - 5} \right)^2} + {\left( {z + 10} \right)^2} = 54\\
2x + z + 1 = y\\
6y - 12 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + 2} \right)^2} + {\left( {y - 5} \right)^2} + {\left( {z + 10} \right)^2} = 54\\
y = 2\\
2x + z + 1 = 2
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + 2} \right)^2} + {\left( {y - 5} \right)^2} + {\left( {z + 10} \right)^2} = 54\\
y = 2\\
z = 1 - 2x
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + 2} \right)^2} + {\left( {2 - 5} \right)^2} + {\left( {1 - 2x + 10} \right)^2} = 54\\
y = 2\\
z = 1 - 2x
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
5{x^2} - 40x + 80 = 0\\
y = 2\\
z = 1 - 2x
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
5{\left( {x - 4} \right)^2} = 0\\
y = 2\\
z = 1 - 2x
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 4\\
y = 2\\
z = - 7
\end{array} \right.
\end{array}$
Vậy hệ có nghiệm duy nhất $\left( {x;y;z} \right) = \left( {4;2; - 7} \right)$
