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If $a \neq p, b \neq q, c \neq r$ and $\left|\begin{array}{ccc}p & b & c \\ p+a & q+b & 2 c \\ a & b & r\end{array}\right|=0$, then $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$ is equal to :
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The correct answer is:
2
We have,
$\left|\begin{array}{ccc}p & b & c \\ p+a & q+b & 2 c \\ a & b & r\end{array}\right|=0$
$\Rightarrow \quad\left|\begin{array}{lll}p & b & c \\ p & b & c \\ a & b & r\end{array}\right|+\left|\begin{array}{lll}p & b & c \\ a & q & c \\ a & b & r\end{array}\right|=0$
$\Rightarrow \quad 0+\left|\begin{array}{lll}p & b & c \\ a & q & c \\ a & b & r\end{array}\right|=0$
$\Rightarrow p(q r-b c)-b(a r-a c)+c(a b-a q)=0$
$\Rightarrow \quad p q r-p b c-b a r+b a c+a b c-a c q=0$
$\Rightarrow \quad p q r-p b c-b a r-a c q=-2 a b c$
$-p q r+p b c+b a r+a c q=2 a b c$
On simplifying $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=2$
$\left|\begin{array}{ccc}p & b & c \\ p+a & q+b & 2 c \\ a & b & r\end{array}\right|=0$
$\Rightarrow \quad\left|\begin{array}{lll}p & b & c \\ p & b & c \\ a & b & r\end{array}\right|+\left|\begin{array}{lll}p & b & c \\ a & q & c \\ a & b & r\end{array}\right|=0$
$\Rightarrow \quad 0+\left|\begin{array}{lll}p & b & c \\ a & q & c \\ a & b & r\end{array}\right|=0$
$\Rightarrow p(q r-b c)-b(a r-a c)+c(a b-a q)=0$
$\Rightarrow \quad p q r-p b c-b a r+b a c+a b c-a c q=0$
$\Rightarrow \quad p q r-p b c-b a r-a c q=-2 a b c$
$-p q r+p b c+b a r+a c q=2 a b c$
On simplifying $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=2$
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