Question

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

Answer 1

Let $∆=\left|\begin{array}{ccc}p& b& c\\ a& q& c\\ a& b& r\end{array}\right|$.
Now,
$$\begin{gathered} ∆=\left|\begin{array}{ccc}p& b& c\\ a& q& c\\ a& b& r\end{array}\right| \\ =\left|\begin{array}{ccc}p& b& c\\ 0& q-b& c-r\\ a& b& r\end{array}\right|\left[\mathrm{Applying}{R}_2\to R_2-R_3\right] \\ =p\left[r\left(q-b\right)-b\left(c-r\right)\right]+a\left[b\left(c-r\right)-c\left(q-b\right)\right]\left[\mathrm{Expanding}\mathrm{along}\mathrm{first}\mathrm{column}\right] \\ =pr\left(q-b\right)+pb\left(r-c\right)-ab\left(r-c\right)-ac\left(q-b\right) \\ =\left(pr-ac\right)\left(q-b\right)+b\left(p-a\right)\left(r-c\right) \\ \mathrm{Since},∆=0. \\ \therefore \left(pr-ac\right)\left(q-b\right)+b\left(p-a\right)\left(r-c\right)=0 \\ \Rightarrow \frac{pr-ac}{\left(p-a\right)\left(r-c\right)}+\frac{b}{q-b}=0 \\ \Rightarrow \frac{pr-ar+ar-ac}{\left(p-a\right)\left(r-c\right)}+\frac{b}{q-b}=0 \\ \Rightarrow \frac{r\left(p-a\right)+a\left(r-c\right)}{\left(p-a\right)\left(r-c\right)}+\frac{b}{q-b}=0 \\ \Rightarrow \frac{r}{r-c}+\frac{a}{p-a}+\frac{b}{q-b}=0 \\ \Rightarrow \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=\frac{p}{p-a}+\frac{q}{q-b}-\frac{a}{p-a}-\frac{b}{q-b} \\ \Rightarrow \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=\frac{p-a}{p-a}+\frac{q-b}{q-b} \\ \Rightarrow \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=2 \\ \mathrm{Hence},\mathrm{the}\mathrm{value}\mathrm{of}\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}\mathrm{is}2. \end{gathered}$$