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

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

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$2$

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(C) Given the determinant equation: $\left|\begin{array}{ccc}p & b & c \ p+a & q+b & 2c \ a & b & r\end{array}ight|=0$.Using the property of determinants,we can 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 $E = \frac{p}{p-a} + \frac{q}{q-b} + \frac{r}{r-c}$.Let $x = p-a, y = q-b, z = r-c$. Then $p = x+a, q = y+b, r = z+c$.Substituting these into the determinant expansion or simplifying the expression directly leads to the value $2$.

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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