\(\begin{array}{l}
7)\,\frac{{bz - cy}}{a} = \frac{{cx - az}}{b} = \frac{{ay - bx}}{c}\\
\Rightarrow \frac{{abz - acy}}{{{a^2}}} = \frac{{bcx - baz}}{{{b^2}}} = \frac{{acy - bcx}}{{{c^2}}} = \frac{{abz - acy + bcx - abz + acy - bcx}}{{{a^2} + {b^2} + {c^2}}} = 0\\
\Rightarrow abz - acy = 0 \Rightarrow bz = cy \Leftrightarrow \frac{y}{b} = \frac{z}{c}\\
bcx - baz = 0 \Rightarrow cx = az \Leftrightarrow \frac{z}{c} = \frac{x}{a}\\
acy - bcx = y0 \Leftrightarrow ay = bx \Leftrightarrow \frac{x}{a} = \frac{y}{b}\\
Suy\,ra\,\,\frac{x}{a} = \frac{y}{b} = \frac{z}{c}\\
8)\,\,A = \left| {\frac{3}{5} - x} \right| + \frac{1}{{2019}} \ge \frac{1}{{2019}}\,\,\left( {do\,\,\left| {\frac{3}{5} - x} \right| \ge 0} \right)\\
{A_{\min }} = \frac{1}{{2019}} \Leftrightarrow \frac{3}{5} - x = 0 \Leftrightarrow x = \frac{3}{5}\\
B = \frac{{2019}}{{2018}} - \left| {x - \frac{9}{5}} \right| \le \frac{{2019}}{{2018}}\,\,\left( {do\,\,\left| {x - \frac{9}{5}} \right| \ge 0} \right)\\
\Rightarrow {B_{\max }} = \frac{{2019}}{{2018}} \Leftrightarrow x - \frac{9}{5} = 0 \Leftrightarrow x = \frac{9}{5}
\end{array}\)
