If $\left|\begin{array}{ccc}a & b-y & c-z \\ a-x & b & c-z \\ a-x & b-y & c\end{array}\right|=0$, then using properties of determinants, find the value of $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}$, where $x, y, z \neq 0$.
$\left|\begin{array}{ccc}a & b-y & c-z \\ a-x & b & c-z \\ a-x & b-y & c\end{array}\right|=0$
$R_{1} \rightarrow R_{1}-R_{2}$
$\left|\begin{array}{ccc}x & -y & 0 \\ a-x & b & c-z \\ a-x & b-y & c\end{array}\right|=0$
$R_{2} \rightarrow R_{2}-R_{3}$
$\Rightarrow\left|\begin{array}{ccc}x & -y & 0 \\ 0 & y & -z \\ a-x & b-y & c\end{array}\right|=0$
Expanding along first row, we get
$x(y c+z b-z y)+y(0-z a+z x)=0$
$\Rightarrow x y c+x z b-x y z+z y a-x y z=0$
$x y c+x z b-2 x y z+z y a=0$
Dividing by $\mathrm{xyz}$, we get
$\frac{c}{z}+\frac{b}{y}-2+\frac{a}{x}=0$
$\therefore \frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2$