If a ne p,b ne q,c ne r and left| {,begin{arr | vedclass

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

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$2$

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(b) $\Delta = \left| {\,\begin{array}{*{20}{c}}p&b&c\{p + a}&{q + b}&{2c}\a&b&r\end{array}\,} ight|\, = 0$ Applying${R_2} 	o {R_2} - {R_1}$

= $\left| {\,\begin{array}{*{20}{c}}{p\,\,\,}&{b\,\,}&c\{a\,\,\,}&{q\,\,}&c\{a\,\,\,}&{b\,\,}&r\end{array}\,} ight|\, = 0$

Applying ${R_2} 	o {R_2} - {R_1}$ and ${R_3} 	o {R_3} - {R_1}$

$\left| {\,\begin{array}{*{20}{c}}p&b&c\{a - p}&{q - b}&0\{a - p}&0&{r - c}\end{array}\,} ight|\, = 0$

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)}}} ight] = 0$

==> $(p - a)(q - b)(r - c)$$\left[ {\frac{p}{{(p - a)}} + \frac{q}{{(q - b)}} - 1 + \frac{r}{{(r - c)}} - 1} ight] = 0$

$	herefore $ $p 
e a,\,q 
e b,\,r 
e c$

$	herefore $ $\frac{p}{{p - a}} + \frac{q}{{q - b}} + \frac{r}{{r - c}} = 2$.

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}} = $

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