Question

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

Answer 1

Let $\Delta =\left|\begin{array}{ccc}p& b& c\\ a& q& c\\ a& b& r\end{array}\right|$
$R_1\to R_1-R_2$ and $R_2\to R_2-R_3$
$\therefore \Delta =\left|\begin{array}{ccc}p-a& b-q& 0\\ 0& q-b& c-r\\ a& b& r\end{array}\right|$
Expanding along $C_1$, we get
$=(p-a)\left\{r\right(q-b)-b(c-r\left)\right\}+a(b-q)(c-r)$
$=r(p-a)(q-b)+b(p-a)(r-c)+a(b-q)(c-r)$
$=(p-a)(q-b)(r-c)\{\frac{r}{r-c}+\frac{b}{q-b}+\frac{a}{p-a}\}$
$\Delta =(p-a)(q-b)(r-c)\{\frac{r}{r-c}+\frac{q}{q-b}-1+\frac{p}{p-a}-1\}$
But given $\Delta =0$,
$\therefore (p-a)(q-b)(r-c)\{\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}-2\}=0$
$\therefore \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}-2=0$ (since $(p-a)(q-b)(r-c)\ne 0$)
Hence $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=2$