If  $\left|\begin{array}{ccc}p& q-b& r-c\\ p-a& q& r-c\\ p-a& q-b & r\end{array}\right|=0$ , then the value of $\frac{\mathrm{p}}{\mathrm{a}}+\frac{\mathrm{q}}{\mathrm{b}}+\frac{\mathrm{r}}{\mathrm{c}}$ is

$=\left|\begin{array}{ccc}p& q-b& r-c\\ p-a& q& r-c\\ p-a& q-b& r\end{array}\right|$

${\mathrm{R}}_1:{\mathrm{R}}_1-{\mathrm{R}}_2$ and ${\mathrm{R}}_2:{\mathrm{R}}_2-{\mathrm{R}}_3$ gives $\left|\begin{array}{ccc}a& -b& 0\\ 0& b& -c\\ p-a& q-b& r\end{array}\right|=0$

Expanding along R1 then , $a(b r+c q-c b)+b(c p-c a)=0$

i.e.$a b r+a c q+b c p=2 a b c$

dividing by $abc$, $\frac{\mathrm{p}}{\mathrm{a}}+\frac{\mathrm{q}}{\mathrm{b}}+\frac{\mathrm{r}}{\mathrm{c}}=2$.