If $a ≠ p,  b ≠ q, c ≠ 0 $ and $\begin{vmatrix} p & b & c \\ p+a & q+b & 2c \\ a & b & r \end {vmatrix}=0,$ then $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$=

2

The correct answer is option (1) : 2

We have,

$\begin{vmatrix} p & b & c \\ p+a & q+b & 2c \\ a & b & r \end {vmatrix}=0$

$⇒\begin{vmatrix} p & b & c \\ a-p & q-b & 0\\ a-p & 0 & r-c \end {vmatrix}=0$        $\begin{bmatrix} Applying\, R_2→R_2-2R_1,\\R_3→R_3-R_1\end{bmatrix}$

$⇒p(q-b)(r-c)-(a-p)b(r-c)-(a-p)c(q-b)=0$

$⇒p(q-b)(r-c)+b(p-a)(r-c)+c(p-a)(q-b)=0$

$⇒\frac{p}{p-a}+\frac{b}{q-b}+\frac{c}{r-c}=0 $    $\begin {bmatrix} \text{Dividing throughout by}\\ (p-a)(q-b)(r-c)\end{bmatrix}$

$⇒\frac{p}{p-a}+\left(\frac{b}{q-b}+1\right) + \left(\frac{c}{r-c}+1\right) = 2 $

$⇒\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}= 2 $