Solve this

$\left|\begin{array}{ccc}a & b-y & c-z \\ a-x & b & c-z \\ a-x & b-y & c\end{array}\right|=0$

$R_{1} \rightarrow R_{1}-R_{2}$

$\left|\begin{array}{ccc}x & -y & 0 \\ a-x & b & c-z \\ a-x & b-y & c\end{array}\right|=0$

$R_{2} \rightarrow R_{2}-R_{3}$

$\Rightarrow\left|\begin{array}{ccc}x & -y & 0 \\ 0 & y & -z \\ a-x & b-y & c\end{array}\right|=0$

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

$\frac{c}{z}+\frac{b}{y}-2+\frac{a}{x}=0$

$\therefore \frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2$