A) \[-1\]
B) \[1\]
C) \[-2\]
D) \[2\]
Correct Answer: D
Solution :
[d] \[Given\left| \begin{matrix} p & b & c \\ a & q & c \\ a & b & r \\ \end{matrix} \right|=0\] \[{{R}_{1}}\to {{R}_{1}}-{{R}_{2}},{{R}_{2}}\to {{R}_{2}}-{{R}_{3}}\] reduces the determinant to \[\left| \begin{matrix} p-a & b-q & 0 \\ 0 & q-b & c-r \\ a & b & r \\ \end{matrix} \right|=0\] \[\Rightarrow (p-a)(q-b)r+a(b-q)(c-r)-b(p-a)(c-r)=0\]\[\Rightarrow \,\] Dividing throughout by \[(p-a)(q-b)(r-c),\] we get \[\Rightarrow \frac{r}{r-c}+\frac{a}{p-a}+\frac{b}{q-b}=0\] \[\Rightarrow \frac{r}{r-c}+\frac{a}{p-a}+\frac{b}{q-b}=2\]You need to login to perform this action.
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