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If $a \ne p,b \ne q,c \ne r$ and $\left| {\,\begin{array}{*{20}{c}}p&b&c\\{p + a}&{q + b}&{2c}\\a&b&r\end{array}\,} \right|$ =$ 0$, then $\frac{p}{{p - a}} + \frac{q}{{q - b}} + \frac{r}{{r - c}} = $
$3$
$2$
$1$
$0$
Solution
(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 $ $\frac{p}{{p – a}} + \frac{q}{{q – b}} + \frac{r}{{r – c}} = 2$.
