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
B
2
C
3
D
c

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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