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Question

If  $$a\neq p,b\neq q,\ c\neq r$$ and $$\begin{vmatrix}
p & b & c\\
 a& q &c \\
 a& b & r
\end{vmatrix}$$ = 0  then the value of  $$\displaystyle \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$$ is equal to


A
1
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B
2
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C
3
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D
c
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Solution

The correct option is A 2
$$\begin{vmatrix} p & b & c \\ a & q & c \\ a & b & r \end{vmatrix}=0$$
$${ R }_{ 1 }\rightarrow { R }_{ 1 }-{ R }_{ 2 },{ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 3 }$$
$$\begin{vmatrix} p-a & b-q & 0 \\ 0 & q-b & c-r \\ a & b & r \end{vmatrix}=0$$
$$(p-a)[r(q-b)-b(c-r)]-(b-q)[-a(c-r)]=0$$
$$r(p-a)(q-b)+b(p-a)(r-c)+a(q-b)(r-c)=0$$
$$\Rightarrow \displaystyle \frac { r }{ r-c } +\frac { b }{ q-b } +\frac { a }{ p-a } =0$$
$$\Rightarrow\displaystyle \frac { r }{ r-c } =-\frac { b }{ q-b } -\frac { a }{ p-a }$$
Now, consider $$\displaystyle \frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$$
$$=\dfrac { p }{ p-a } +\dfrac { q }{ q-b } -\dfrac { b }{ q-b } -\dfrac { a }{ p-a } $$
$$=1+1=2$$

Mathematics

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