A) \[0\]
B) \[1\]
C) \[-1\]
D) \[2\]
Correct Answer: D
Solution :
We have,\[\left| \begin{matrix} p & b & c \\ a & q & c \\ a & b & r \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[\left| \begin{matrix} p & b & c \\ a-p & q-b & 0 \\ 0 & b-q & r-c \\ \end{matrix} \right|=0\] [Applying\[{{R}_{3}}\to {{R}_{3}}-{{R}_{2}}\]and\[{{R}_{2}}\to {{R}_{2}}-{{R}_{1}}]\] \[\Rightarrow \] \[\left| \begin{matrix} \frac{p}{p-a} & \frac{b}{q-b} & \frac{c}{r-c} \\ -1 & 1 & 0 \\ 0 & -1 & 1 \\ \end{matrix} \right|=0\] \[\Rightarrow \] \[\frac{p}{p-a}+\frac{b}{q-b}+\frac{c}{r-c}=0\] \[\Rightarrow \] \[\frac{p}{p-a}+\left( \frac{q}{q-b}-1 \right)+\left( \frac{r}{r-c}-1 \right)=0\] \[\Rightarrow \] \[\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=0\]
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