If $a+x=b+y = c+ z+1, $ where $a,b,c,x,y,z $ are non-zero distinct real numbers, then $\begin{vmatrix}x& a+y & x+a\\y& b+y & y+b\\z& c+y & z+c\end{vmatrix} $ is equal to

$y(a-b)$

The correct answer is option (1) : $y(a-b)$

$\begin{vmatrix}x& a+y & x+a\\y& b+y & y+b\\z& c+y & z+c\end{vmatrix} $

$=\begin{vmatrix}x& a+y & a\\y& b+y & b\\z& c+y & c\end{vmatrix} $ Applying $C_3→C_2-C_1$

$=\begin{vmatrix}x& y & a\\y& y & b\\z& y & c\end{vmatrix} $ Applying $C_2→C_2-C_3$

$= y\begin{vmatrix}x& 1 & a\\y& 1& b\\z& 1 & c\end{vmatrix} $