If $a \neq p, b \neq q, c \neq r$ and $\left|\begin{array}{lll}p & b & c \\ a & q & c \\ a & b & r\end{array}\right|=0 \quad$ then the value of $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$ is :

2

By operating $R_1 \rightarrow R_1-R_2$ and $R_2 \rightarrow R_2-R_3$

$0=\left|\begin{array}{ccc} p-a & b-q & 0 \\ 0 & q-b & c-r \\ a & b & r \end{array}\right|$

$=(p-a)\{r(q-b)-b(c-r)\}+a(b-q)(c-r)$

$=(p-a)(r q-r b)+a(b-q)(c-r)+b(p-a)(r-c)$

dividing throughout by $(p-a)(q-b)(r-c)$,

$=\frac{r}{r-c}+\frac{b}{q-b}+\frac{a}{p-a} \Rightarrow \frac{r}{r-c}+\left(\frac{b}{q-b}+1\right)+\left(\frac{a}{p-a}+1\right)=2$

Therefore $\frac{r}{r-c}+\frac{q}{q-b}+\frac{p}{p-a}=2$.

Hence (3) is the correct answer.