If det [ p b c

a q c = 0 then find (p/p-a) + (q/q-b) + (r/r-c)

a b r]

the given determinant is:
$\left|\begin{array}{c}p\\ a\\ a\end{array}\begin{array}{c}b\\ q\\ b\end{array}\begin{array}{c}c\\ c\\ r\end{array}\right|=0$
​$$\begin{gathered} applying R_1\to R_1-R_3 and R_2\to R_2-R_3 \\ \left|\begin{array}{c}p-a\\ 0\\ a\end{array}\begin{array}{c}0\\ q-b\\ b\end{array}\begin{array}{c}c-r\\ c-r\\ r\end{array}\right|=0 \\ taking common (p-a) from C_1 ; (q-b) from C_2 and (c-r) from C_3 \\ (p-a)(q-b)(c-r)\left|\begin{array}{c}1\\ 0\\ \frac{a}{p-a}\end{array}\begin{array}{c}0\\ 1\\ \frac{b}{q-b}\end{array}\begin{array}{c}1\\ 1\\ \frac{r}{c-r}\end{array}\right|=0 \end{gathered}$$
since p, q, r and a, b , c are different numbers
$\left|\begin{array}{c}1\\ 0\\ \frac{a}{p-a}\end{array}\begin{array}{c}0\\ 1\\ \frac{b}{q-b}\end{array}\begin{array}{c}1\\ 1\\ \frac{r}{c-r}\end{array}\right|=0$
expanding the determinant:
$$\begin{gathered} \frac{r}{c-r}-\frac{b}{q-b}+0+1*\left[0-\frac{a}{p-a}\right]=0 \\ \frac{r}{c-r}-\frac{b}{q-b}-\frac{a}{p-a}=0 \\ -\frac{r}{r-c}-\frac{b}{q-b}-\frac{a}{p-a}=0 \\ \Rightarrow \frac{r}{r-c}+\frac{b}{q-b}+\frac{a}{p-a}=0 \end{gathered}$$
adding 2 on the both sides:
$$\begin{gathered} \frac{r}{r-c}+\frac{b}{q-b}+1+\frac{a}{p-a}+1=2 \\ \frac{r}{r-c}+\frac{b+q-b}{q-b}+\frac{a+p-a}{p-a}=2 \\ \frac{r}{r-c}+\frac{q}{q-b}+\frac{p}{p-a}=2 \end{gathered}$$

hope this helps you