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

$\begin{array}{1 1}(A)\;3&(B)\;2\\(C)\;1&(D)\;0\end{array}$

$\begin{vmatrix}p&b&c\\p+a&q+b&2c\\a&b&r\end{vmatrix}=0$
Applying $R_2=R_2-R_1$ we get
$\begin{vmatrix}p&b&c\\a&p&c\\a&b&r\end{vmatrix}=0$
Applying $R_1=R_1-R_2,R_2=R_2-R_3$ we get
$\begin{vmatrix}p-a&b-q&0\\0&q-b&c-r\\a&b&r\end{vmatrix}=0$
Applying $\large\frac{1}{p-a}$$C_1,\large\frac{1}{q-b}$$C_2,\large\frac{1}{r-c}$$C_3 \begin{vmatrix}1&-1&0\\0&1&-1\\\large\frac{a}{p-a}&\large\frac{b}{q-b}&\large\frac{r}{r-c}\end{vmatrix}=0 Applying C_1+C_2+C_3 and expanding with C_1 we get \large\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}=0 \Rightarrow (\large\frac{a}{p-a}+$$1)+(\large\frac{b}{q-b}$$+1)+\large\frac{r}{r-c}$$=1+1$
$\Rightarrow\large\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$=2
Hence (B) is the correct answer.