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:

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Determinants

a + x = b + y = c + z + 1

, where

a, b, c, x, y, z

are non-zero distinct real numbers then

x y z a + y b + y c + y x + a y + b z + c

is equal to:

Answer: (A)

Starting from

x y z a + y b + y c + y x + a y + b z + c

, apply

C_{3} \to C_{3} - C_{1}

, which turns the third column into

(a, b, c)

. Then apply

C_{2} \to C_{2} - C_{3}

, turning the second column into

(y, y, y)

, and factor

y

out to get

y x y z 111 a b c

. Expanding this determinant gives

x c - x b - y c + b z + a y - a z

, and substituting the given condition

a + x = b + y = c + z + 1

collapses it to

a - b

. The determinant is therefore

y (a - b)

JEE Main · 2020 · 05 Sep 2020 · Shift 2 · Single correct option