If a ne p, b ne q, c ne r and begin{vmatrix} | vedclass

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(B) Given the determinant $\Delta = \begin{vmatrix} p & b & c \ p + a & q + b & 2c \ a & b & r \end{vmatrix} = 0$. Applying $R_2 	o R_2 - R_1$,we g

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$2$

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(B) Given the determinant $\Delta = \begin{vmatrix} p & b & c \ p + a & q + b & 2c \ a & b & r \end{vmatrix} = 0$. Applying $R_2 	o R_2 - R_1$,we get: $\begin{vmatrix} p & b & c \ a & q & c \ a & b & r \end{vmatrix} = 0$. Now,applying $R_2 	o R_2 - R_3$ and $R_1 	o R_1 - R_3$,we get: $\begin{vmatrix} p - a & 0 & c - r \ 0 & q - b & c - r \ a & b & r \end{vmatrix} = 0$. Expanding the determinant: $(p - a)[(q - b)r - b(c - r)] + (c - r)[0 - a(q - b)] = 0$. $(p - a)(q - b)r - (p - a)b(c - r) - a(q - b)(c - r) = 0$. Dividing by $(p - a)(q - b)(r - c)$ (since $p 
e a, q 
e b, r 
e c$): $\frac{r}{r - c} + \frac{b}{q - b} + \frac{a}{p - a} = 0$. Using $\frac{x}{x - y} = 1 + \frac{y}{x - y}$,we rewrite the terms: $(1 + \frac{p}{p - a} - 1) + (1 + \frac{q}{q - b} - 1) + (1 + \frac{r}{r - c} - 1) = 0$ is not the direct path. Actually,$\frac{p}{p - a} = 1 + \frac{a}{p - a}$. Thus,$\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} = (1 + \frac{a}{p - a}) + (1 + \frac{b}{q - b}) + (1 + \frac{c}{r - c}) = 3 + (\frac{a}{p - a} + \frac{b}{q - b} + \frac{c}{r - c})$. From the expansion,we found $\frac{a}{p - a} + \frac{b}{q - b} + \frac{c}{r - c} = -1$. Therefore,$3 - 1 = 2$.

If $a \ne p, b \ne q, c \ne r$ and $\begin{vmatrix} p & b & c \\ p + a & q + b & 2c \\ a & b & r \end{vmatrix} = 0$,then $\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} = $

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