Đáp án:

 1) $\left( {x;y;z} \right) = \left( {9;18;27} \right)$

2) $\left( {x;y;z} \right) \in \left\{ {\left( {\dfrac{{6\sqrt {93} }}{{31}};\dfrac{{9\sqrt {93} }}{{31}};\dfrac{{10\sqrt {93} }}{{31}}} \right),\left( {\dfrac{{ - 6\sqrt {93} }}{{31}};\dfrac{{ - 9\sqrt {93} }}{{31}};\dfrac{{ - 10\sqrt {93} }}{{31}}} \right)} \right\}$

3) $\left( {x;y;z} \right) = \left( {75;50;30} \right)$

Giải thích các bước giải:

$\begin{array}{l}
1)x = \dfrac{y}{2} = \dfrac{z}{3} \Rightarrow \dfrac{x}{1} = \dfrac{y}{2} = \dfrac{z}{3} = \dfrac{{4x - 3y + 2z}}{{4.1 - 3.2 + 2.3}} = \dfrac{{36}}{4} = 9\\
 \Rightarrow \left\{ \begin{array}{l}
\dfrac{x}{1} = 9 \Rightarrow x = 9\\
\dfrac{y}{2} = 9 \Rightarrow y = 18\\
\dfrac{z}{3} = 9 \Rightarrow z = 27
\end{array} \right.
\end{array}$

Vậy $\left( {x;y;z} \right) = \left( {9;18;27} \right)$

$\begin{array}{l}
2)\dfrac{x}{y} = \dfrac{2}{3};\dfrac{x}{z} = \dfrac{3}{5}\\
 \Leftrightarrow \dfrac{x}{2} = \dfrac{y}{3};\dfrac{x}{3} = \dfrac{z}{5}\\
 \Leftrightarrow \dfrac{x}{6} = \dfrac{y}{9};\dfrac{x}{6} = \dfrac{z}{{10}}\\
 \Leftrightarrow \dfrac{x}{6} = \dfrac{y}{9} = \dfrac{z}{{10}}\left( 1 \right)\\
 \Rightarrow {\left( {\dfrac{x}{6}} \right)^2} = {\left( {\dfrac{y}{9}} \right)^2} = {\left( {\dfrac{z}{{10}}} \right)^2} = \dfrac{{{x^2} + {y^2} + {z^2}}}{{{6^2} + {9^2} + {{10}^2}}} = \dfrac{{21}}{{217}} = \dfrac{3}{{31}}\\
 \Rightarrow \dfrac{{{x^2}}}{{36}} = \dfrac{3}{{31}} \Rightarrow {x^2} = \dfrac{{108}}{{31}} \Rightarrow \left[ \begin{array}{l}
x = \dfrac{{6\sqrt {93} }}{{31}}\\
x = \dfrac{{ - 6\sqrt {93} }}{{31}}
\end{array} \right.\\
 + )x = \dfrac{{6\sqrt {93} }}{{31}}\mathop  \Rightarrow \limits^{\left( 1 \right)} \dfrac{y}{9} = \dfrac{z}{{10}} = \dfrac{{\dfrac{{6\sqrt {93} }}{{31}}}}{6} = \dfrac{{\sqrt {93} }}{{31}} \Rightarrow \left\{ \begin{array}{l}
y = \dfrac{{9\sqrt {93} }}{{31}}\\
z = \dfrac{{10\sqrt {93} }}{{31}}
\end{array} \right.\\
 + )x = \dfrac{{ - 6\sqrt {93} }}{{31}}\mathop  \Rightarrow \limits^{\left( 1 \right)} \dfrac{y}{9} = \dfrac{z}{{10}} = \dfrac{{\dfrac{{ - 6\sqrt {93} }}{{31}}}}{6} = \dfrac{{ - \sqrt {93} }}{{31}} \Rightarrow \left\{ \begin{array}{l}
y = \dfrac{{ - 9\sqrt {93} }}{{31}}\\
z = \dfrac{{ - 10\sqrt {93} }}{{31}}
\end{array} \right.
\end{array}$

Vậy $\left( {x;y;z} \right) \in \left\{ {\left( {\dfrac{{6\sqrt {93} }}{{31}};\dfrac{{9\sqrt {93} }}{{31}};\dfrac{{10\sqrt {93} }}{{31}}} \right),\left( {\dfrac{{ - 6\sqrt {93} }}{{31}};\dfrac{{ - 9\sqrt {93} }}{{31}};\dfrac{{ - 10\sqrt {93} }}{{31}}} \right)} \right\}$

$\begin{array}{l}
3)2x = 3y = 5z\\
 \Leftrightarrow \dfrac{{2x}}{{30}} = \dfrac{{3y}}{{30}} = \dfrac{{5z}}{{30}}\\
 \Leftrightarrow \dfrac{x}{{15}} = \dfrac{y}{{10}} = \dfrac{z}{6}\\
 \Rightarrow \dfrac{x}{{15}} = \dfrac{y}{{10}} = \dfrac{z}{6} = \dfrac{{x + y - z}}{{15 + 10 - 6}} = \dfrac{{95}}{{19}} = 5\\
 \Rightarrow \left\{ \begin{array}{l}
\dfrac{x}{{15}} = 5 \Rightarrow x = 75\\
\dfrac{y}{{10}} = 5 \Rightarrow y = 50\\
\dfrac{z}{6} = 5 \Rightarrow z = 30
\end{array} \right.
\end{array}$

Vậy $\left( {x;y;z} \right) = \left( {75;50;30} \right)$