Đáp án:

$\begin{array}{l}
a)x:y:z = 10:3:5\\
 \Leftrightarrow \dfrac{x}{{10}} = \dfrac{y}{3} = \dfrac{z}{5} = \dfrac{{x + y + z}}{{10 + 3 + 5}} = \dfrac{{0,18}}{{18}} = \dfrac{1}{{100}}\\
 \Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{1}{{10}}\\
y = \dfrac{3}{{100}}\\
z = \dfrac{1}{{20}}
\end{array} \right.\\
Vậy\,x = \dfrac{1}{{10}};y = \dfrac{3}{{100}};z = \dfrac{1}{{20}}\\
b)x:y:z = 10:3:5\\
 \Leftrightarrow \dfrac{x}{{10}} = \dfrac{y}{3} = \dfrac{z}{5}\\
 = \dfrac{{3x}}{{30}} = \dfrac{{2y}}{6} = \dfrac{{4z}}{{20}}\\
 = \dfrac{{3x + 2y - 4z}}{{30 + 6 - 20}} = \dfrac{{ - 1,6}}{{16}} =  - \dfrac{1}{{10}}\\
 \Leftrightarrow \left\{ \begin{array}{l}
x =  - 1\\
y =  - \dfrac{3}{{10}}\\
z = \dfrac{{ - 1}}{2}
\end{array} \right.\\
Vậy\,x =  - 1;y =  - \dfrac{3}{{10}};z =  - \dfrac{1}{2}\\
c)\dfrac{x}{{10}} = \dfrac{y}{3} = \dfrac{z}{5} = k \Leftrightarrow \left\{ \begin{array}{l}
x = 10k\\
y = 3k\\
z = 5k
\end{array} \right.\\
x.y.z =  - 1200\\
 \Leftrightarrow 10k.3k.5k =  - 1200\\
 \Leftrightarrow {k^3} =  - 8\\
 \Leftrightarrow k =  - 2\\
 \Leftrightarrow \left\{ \begin{array}{l}
x =  - 20\\
y =  - 6\\
z =  - 10
\end{array} \right.\\
Vậy\,x =  - 20;y =  - 6;z =  - 10\\
2)\left\{ \begin{array}{l}
4x = 5y\\
6y = 7z
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\dfrac{x}{5} = \dfrac{y}{4} \Leftrightarrow \dfrac{x}{{35}} = \dfrac{y}{{28}}\\
\dfrac{y}{7} = \dfrac{z}{6} \Leftrightarrow \dfrac{y}{{28}} = \dfrac{z}{{24}}
\end{array} \right.\\
 \Leftrightarrow \dfrac{x}{{35}} = \dfrac{y}{{28}} = \dfrac{z}{{24}} = \dfrac{{z - x}}{{24 - 35}} = \dfrac{{ - 11}}{{ - 11}} = 1\\
 \Leftrightarrow \left\{ \begin{array}{l}
x = 35\\
y = 28\\
z = 24
\end{array} \right.\\
Vậy\,x = 35;y = 28;z = 24\\
3)\\
\dfrac{{3 + 2y}}{3} = \dfrac{{3 - 4y}}{{4x}} = \dfrac{{3 - 10y}}{{3x}}\\
 \Leftrightarrow 3x\left( {3 - 4y} \right) = 4x\left( {3 - 10y} \right)\\
 \Leftrightarrow 9 - 12y = 12 - 40y\left( {do:x\# 0} \right)\\
 \Leftrightarrow 28y = 3\\
 \Leftrightarrow y = \dfrac{3}{{28}}\\
 \Leftrightarrow \dfrac{{3 + 2.\dfrac{3}{{28}}}}{3} = \dfrac{{3 - 4.\dfrac{3}{{28}}}}{{4x}}\\
 \Leftrightarrow x = \dfrac{3}{5}\\
Vậy\,x = \dfrac{3}{5};y = \dfrac{3}{{28}}
\end{array}$