If `x!=0,y!=0,z!=0a n d|1+x1 1 1+y1+2y1 1+z1+z1+3z|=0` , then `x^(-1)+y^(-1)+z^(-1)` is equal to `1` b. `-1` c. `-3` d. none of these

Options

A. `0`
B. `1`
C. `3`
D. `6`

Answer

C

Explanation

`(c )` `Delta=|{:(1+x,1,1),(1+y,1+2y,1),(1+z,1+z,1+3z):}|=0`
Applying `C_(1)toC_(1)-C_(3)`, `C_(2)toC_(2)-C_(1)` we get
`Delta=|{:(x,0,1),(y,2y,0),(-2z,-2z,1+3z):}|=0`
`:.Delta=|{:(x,0,0),(y,2y,1-(y)/(x)),(-2z,-2z,1+3z+(2z)/(x)):}|=0` (by `C_(2)toC_(3)-(1)/(x)C_(1)`)
`:.x[2y(1+3z+(2z)/(x))+2z(1-(y)/(x))]=0`
`:.[2xy+6zxy+4yz+2zx-2yz]=0`
`:.2(xyz)[(1)/(x)+(1)/(y)+(1)/(z)+3]=0`
`=-((1)/(x)+(1)/(y)+(1)/(z))=3`