= $\left| {\,\begin{array}{*{20}{c}}{p\,\,\,}&{b\,\,}&c\\{a\,\,\,}&{q\,\,}&c\\{a\,\,\,}&{b\,\,}&r\end{array}\,} \right|\, = 0$

Applying ${R_2} \to {R_2} - {R_1}$ and ${R_3} \to {R_3} - {R_1}$

$\left| {\,\begin{array}{*{20}{c}}p&b&c\\{a - p}&{q - b}&0\\{a - p}&0&{r - c}\end{array}\,} \right|\, = 0$

On expansion we get,

$p\,(q - b)(r - c) - b(a - p)(r - c) - c(q - b)(a - p) = 0$

==> $(p - a)(q - b)(r - c)$$\left[ {\frac{p}{{(p - a)}} + \frac{b}{{(q - b)}} + \frac{c}{{(r - c)}}} \right] = 0$

==> $(p - a)(q - b)(r - c)$$\left[ {\frac{p}{{(p - a)}} + \frac{q}{{(q - b)}} - 1 + \frac{r}{{(r - c)}} - 1} \right] = 0$

$\therefore $ $p \ne a,\,q \ne b,\,r \ne c$

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