Đáp án:

$\begin{array}{l}
 + )xz = {y^2} \Rightarrow 2xz = 2{y^2}\\
 + ){x^2} + {z^2} + 99 = 7{y^2}\\
 \Rightarrow {x^2} + {z^2} + 2xz + 99 = 7{y^2} + 2{y^2}\\
 \Rightarrow {\left( {x + z} \right)^2} + 99 = 9{y^2} = {\left( {3y} \right)^2}\\
 \Rightarrow {\left( {x + z} \right)^2} - {\left( {3y} \right)^2} =  - 99\\
 \Rightarrow \left( {x + z + 3y} \right)\left( {x + z - 3y} \right) =  - 99 =  - \left( {9.11} \right) =  - \left( {3.33} \right) =  - \left( {99.1} \right)\\
Goi\,x + z = a;\,3y = b\\
 \Rightarrow \left( {a + b} \right)\left( {a - b} \right) =  - \left( {9.11} \right) =  - \left( {3.33} \right) =  - \left( {99.1} \right)\\
 + TH1:\left( {a + b} \right)\left( {a - b} \right) =  - \left( {9.11} \right)\\
 \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
a + b = 11\\
a - b =  - 9
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a = 1\\
b = 10
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x + z = 1\\
3y = 10
\end{array} \right.\left( {ktm\,} \right)\\
\left\{ \begin{array}{l}
a + b = 9\\
a - b =  - 11
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a =  - 1\\
b = 10
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x + z =  - 1\\
3y = 10
\end{array} \right.\left( {ktm\,} \right)
\end{array} \right.\\
 + TH2:\left( {a + b} \right)\left( {a - b} \right) =  - \left( {3.33} \right)\\
 \Rightarrow \left[ \begin{array}{l}
\left\{ \begin{array}{l}
a + b = 33\\
a - b =  - 3
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a = 15\\
b = 18
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x + z = 15\\
3y = 18
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x + z = 15\\
y = 6 \Rightarrow xz = {6^2} = 36
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = 12\\
z = 3\\
y = 6
\end{array} \right.\left( {tm} \right)\\
\left\{ \begin{array}{l}
a + b = 3\\
a - b =  - 33
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
a =  - 15\\
b = 18
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x + z =  - 15\\
3y = 18
\end{array} \right.\left( {ktm\,} \right)
\end{array} \right.\\
 + TH3:không\,thỏa\,mãn\\
Vay\,x = 12;y = 6;z = 3\,Hoặc\,x = 3;y = 6;z = 12
\end{array}$