# 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:
$\left|\begin{array}{c}p\\ a\\ a\end{array}\begin{array}{c}b\\ q\\ b\end{array}\begin{array}{c}c\\ c\\ r\end{array}\right|=0$

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:
$\frac{r}{c-r}-\frac{b}{q-b}+0+1*\left[0-\frac{a}{p-a}\right]=0\phantom{\rule{0ex}{0ex}}\frac{r}{c-r}-\frac{b}{q-b}-\frac{a}{p-a}=0\phantom{\rule{0ex}{0ex}}-\frac{r}{r-c}-\frac{b}{q-b}-\frac{a}{p-a}=0\phantom{\rule{0ex}{0ex}}⇒\frac{r}{r-c}+\frac{b}{q-b}+\frac{a}{p-a}=0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$
adding 2 on the both sides:
$\frac{r}{r-c}+\frac{b}{q-b}+1+\frac{a}{p-a}+1=2\phantom{\rule{0ex}{0ex}}\frac{r}{r-c}+\frac{b+q-b}{q-b}+\frac{a+p-a}{p-a}=2\phantom{\rule{0ex}{0ex}}\frac{r}{r-c}+\frac{q}{q-b}+\frac{p}{p-a}=2$

hope this helps you

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