If $a+x=b+y=c+z+1$, where $a, b, c, x, y, z$ are non-zero
distinct real numbers, then $\left|\begin{array}{lll}x & a+y & x+a \\ y & b+y & y+b \\ z & c+y & z+c\end{array}\right|$ is equal to :
Correct Option: 2,
(2) Use properties of determinant
$\left|\begin{array}{lll}x & a+y & x+a \\ y & b+y & y+b \\ z & c+y & z+c\end{array}\right|=\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|+y\left|\begin{array}{ccc}x & 1 & x+a \\ y & 1 & y+b \\ z & 1 & z+c\end{array}\right|$
$=0+y\left|\begin{array}{ccc}x & 1 & x+a \\ y-x & 0 & 0 \\ z-x & 0 & -1\end{array}\right| \quad\left[\begin{array}{l}R_{2} \rightarrow R_{2}-R_{1}, \\ R_{3} \rightarrow R_{3}-R_{1}\end{array}\right]$
$=-y(x-y)=-y(b-a)=y(a-b)$
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