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प्रश्न
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उत्तर
Let
Now,
\[∆ = \begin{vmatrix}p & b & c \\ a & q & c \\ a & b & r\end{vmatrix}\]
\[ = \begin{vmatrix}p & b & c \\ 0 & q - b & c - r \\ a & b & r\end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_3 \right]\]
\[ = p\left[ r\left( q - b \right) - b\left( c - r \right) \right] + a\left[ b\left( c - r \right) - c\left( q - b \right) \right] \left[\text{ Expanding along first column }\right]\]
\[ = pr\left( q - b \right) + pb\left( r - c \right) - ab\left( r - c \right) - ac\left( q - b \right)\]
\[ = \left( pr - ac \right)\left( q - b \right) + b\left( p - a \right)\left( r - c \right)\]
\[\text{ Since, }∆ = 0 . \]
\[ \therefore \left( pr - ac \right)\left( q - b \right) + b\left( p - a \right)\left( r - c \right) = 0\]
\[ \Rightarrow \frac{pr - ac}{\left( p - a \right)\left( r - c \right)} + \frac{b}{q - b} = 0\]
\[ \Rightarrow \frac{pr - ar + ar - ac}{\left( p - a \right)\left( r - c \right)} + \frac{b}{q - b} = 0\]
\[ \Rightarrow \frac{r\left( p - a \right) + a\left( r - c \right)}{\left( p - a \right)\left( r - c \right)} + \frac{b}{q - b} = 0\]
\[ \Rightarrow \frac{r}{r - c} + \frac{a}{p - a} + \frac{b}{q - b} = 0\]
\[ \Rightarrow \frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} = \frac{p}{p - a} + \frac{q}{q - b} - \frac{a}{p - a} - \frac{b}{q - b}\]
\[ \Rightarrow \frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} = \frac{p - a}{p - a} + \frac{q - b}{q - b}\]
\[ \Rightarrow \frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} = 2\]
\[\text{Hence, the value of }\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}\text{ is }2 .\]
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