(b) $\Delta = \left| {\,\begin{array}{*{20}{c}}p&b&c\\{p + a}&{q + b}&{2c}\\a&b&r\end{array}\,} \right|\, = 0$ Applying${R_2} \to {R_2} – {R_1}$

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

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