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 the value of $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$ is equal to

• A. -1
• B. 1
• C. -2
• D. 2

Answer 1

Answer: (D)

Given
$\left|\begin{array}{ccc}p& b& c\\ a& q& c\\ a& b& r\end{array}\right|=0$
$R_1\to R_1-R_2,R_2\to R_2-R_3$ reduces the determinant to $\left|\begin{array}{ccc}p-a& b-q& 0\\ 0& q-b& c-r\\ a& b& r\end{array}\right|=0$ $\Rightarrow (p-a)(q-b)r+a(b-q)(c-r)-b(p-a)(c-r)=0$
$\Rightarrow$ Dividing throughout by $(p-a)(q-b)(r-c)$, we get
$\frac{r}{r-c}+\frac{a}{p-a}+\frac{b}{q-b}=0$
$\Rightarrow \frac{r}{r-c}+1+\frac{a}{p-a}+1+\frac{b}{q-b}=2$
$\Rightarrow \frac{r}{r-c}+\frac{p}{p-a}+\frac{q}{q-b}=2$