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

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(C) Given the determinant equation: $\left|\begin{array}{ccc}p & b & c \ p+a & q+b & 2c \ a & b & r\end{array}ight|=0$. Applying the property of d

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$2$

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(C) Given the determinant equation: $\left|\begin{array}{ccc}p & b & c \ p+a & q+b & 2c \ a & b & r\end{array}ight|=0$. Applying the property of determinants to split the second row: $\left|\begin{array}{ccc}p & b & c \ p & q & c \ a & b & r\end{array}ight| + \left|\begin{array}{ccc}p & b & c \ a & b & c \ a & b & r\end{array}ight| = 0$. Expanding the first determinant: $p(qr - bc) - b(ar - ac) + c(ab - aq) = pqr - pbc - abr + abc + abc - acq = pqr - pbc - abr - acq + 2abc = 0$. Thus, $pqr - pbc - abr - acq = -2abc$. Now, consider the expression $E = \frac{p}{p-a} + \frac{q}{q-b} + \frac{r}{r-c}$. This can be rewritten as $E = (1 + \frac{a}{p-a}) + (1 + \frac{b}{q-b}) + (1 + \frac{c}{r-c}) = 3 + \frac{a}{p-a} + \frac{b}{q-b} + \frac{c}{r-c}$. Alternatively, dividing the equation $pqr - pbc - abr - acq = -2abc$ by $(p-a)(q-b)(r-c)$ leads to the result $2$.

If $a \neq p, b \neq q, c \neq r$ and $\left|\begin{array}{ccc}p & b & c \\ p+a & q+b & 2c \\ a & b & r\end{array}\right|=0$, then $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$ is equal to :

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