Question

\[If \begin{vmatrix}p & b & c \\ a & q & c \\ a & b & r\end{vmatrix} = 0,\text{ find the value of }\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}, p \neq a, q \neq b, r \neq c .\]

 

Answer 1

Let 

\[∆ = \begin{vmatrix}p & b & c \\ a & q & c \\ a & b & r\end{vmatrix}\]

Now,

\[∆ = \begin{vmatrix}p & b & c \\ a & q & c \\ a & b & r\end{vmatrix}\] 

\[ = \begin{vmatrix}p & b & c \\ 0 & q - b & c - r \\ a & b & r\end{vmatrix} \left[\text{ Applying }R_2 \to R_2 - R_3 \right]\] 

\[ = p\left[ r\left( q - b \right) - b\left( c - r \right) \right] + a\left[ b\left( c - r \right) - c\left( q - b \right) \right] \left[\text{ Expanding along first column }\right]\] 

\[ = pr\left( q - b \right) + pb\left( r - c \right) - ab\left( r - c \right) - ac\left( q - b \right)\] 

\[ = \left( pr - ac \right)\left( q - b \right) + b\left( p - a \right)\left( r - c \right)\] 

\[\text{ Since, }∆ = 0 . \] 

\[ \therefore \left( pr - ac \right)\left( q - b \right) + b\left( p - a \right)\left( r - c \right) = 0\] 

\[ \Rightarrow \frac{pr - ac}{\left( p - a \right)\left( r - c \right)} + \frac{b}{q - b} = 0\] 

\[ \Rightarrow \frac{pr - ar + ar - ac}{\left( p - a \right)\left( r - c \right)} + \frac{b}{q - b} = 0\] 

\[ \Rightarrow \frac{r\left( p - a \right) + a\left( r - c \right)}{\left( p - a \right)\left( r - c \right)} + \frac{b}{q - b} = 0\] 

\[ \Rightarrow \frac{r}{r - c} + \frac{a}{p - a} + \frac{b}{q - b} = 0\] 

\[ \Rightarrow \frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} = \frac{p}{p - a} + \frac{q}{q - b} - \frac{a}{p - a} - \frac{b}{q - b}\] 

\[ \Rightarrow \frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} = \frac{p - a}{p - a} + \frac{q - b}{q - b}\] 

\[ \Rightarrow \frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c} = 2\] 

\[\text{Hence, the value of }\frac{p}{p - a} + \frac{q}{q - b} + \frac{r}{r - c}\text{ is }2 .\]