If x,y, z are in A.P., then the values of the determinant `|(a +2,a +3,a + 2y),(a + 3,a +4,a + 2y),(a +4,a +5,a + 2z)|`, is

Options

A. 1
B. 0
C. 2a
D. a

Answer

B

Explanation

since x,y,z are in A.P. therefore `x+z-2y=0` Now
`|{:(a+2,,a+3,,a+2x),(a+3,,a+4,,a+2y),(a+4,,a+5,,a+2x):}|=|{:(0,,0,,2(x+z-2y)),(a+3,,a+4,,a+2y),(a+4,,a+5,,a+2z):}|`
Applying `R_(1) to R_(1) +R_(3) -2R_(2)`
`=|{:(0,,0,,0),(a+3,,a+4,,a+2y),(a+4,,a+5,,a+2z):}|" "[ :' x +z -2y =0]`
`=0`