If $\begin{vmatrix}p & q-y & r-z\\ p-x & q & r-z\\ p-x & q-y & r\end{vmatrix}=0, $ then the value of $\frac{p}{x}+\frac{q}{y}+\frac{r}{z} $is

2

The correct answer is option (4) : 2

Applying $R_1→R_1-R_2, R_2 → R_2 -R_3, $ we have

$\begin{vmatrix} x& -y & 0 \\ 0 & y & -z \\ p-x & q-y & r \end{vmatrix} = 0 $

$⇒x(yr+qz-yz) + y ( pz-xz) = 0 $

$⇒pyz + qzx + rxy = 2xyz ⇒\frac{p}{x}+\frac{q}{y}+\frac{r}{z} = 2$