If $$a \ne p,b \ne q,c \ne r$$    and \[\left| {\begin{array}{*{20}{c}} p&b&c\\ a&q&c\\ a&b&r \end{array}} \right| = 0\]    then the value of $$\frac{p}{{p - a}} + \frac{q}{{q - b}} + \frac{r}{{r - c}}$$     is equal to

Solution :
Given \[\left| {\begin{array}{*{20}{c}} p&b&c\\ a&q&c\\ a&b&r \end{array}} \right| = 0\]
$${R_1} \to {R_1} - {R_2},{R_2} \to {R_2} - {R_3}$$      reduces the determinant to
\[\left| {\begin{array}{*{20}{c}} {p - a}&{b - q}&0\\ 0&{q - b}&{c - r}\\ a&b&r \end{array}} \right| = 0\]
$$ \Rightarrow \,\left( {p - a} \right)\left( {q - b} \right)r + a\left( {b - q} \right)\left( {c - r} \right) - b\left( {p - a} \right)\left( {c - r} \right) = 0$$
$$ \Rightarrow $$ Dividing throughout by $$\left( {p - a} \right)\left( {q - b} \right)\left( {r - c} \right),$$     we get
$$\eqalign{ & \Rightarrow \,\frac{r}{{r - c}} + \frac{a}{{p - a}} + \frac{b}{{q - b}} = 0 \cr & \Rightarrow \,\frac{r}{{r - c}} + \frac{a}{{p - a}} + \frac{b}{{q - b}} = 2 \cr} $$