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The points A(2-x,2,2), B(2,2-y,2), C(2,2...

The points `A(2-x,2,2), B(2,2-y,2), C(2,2,2-z)` and `D(1,1,1)` are coplanar, then locus of `P(x,y,z)` is

A

`1/x+1/y+1/z=1`

B

`x+y+z=1`

C

`1/(1-x)+1/(1-y)+1/(1-z)=1`

D

`1/(1-x)+1/(1-y)+1/(1-z)=2`

Text Solution

Verified by Experts

The correct Answer is:
A
`vec(AB)=vec(OB)-vec(OA)={2hati+(2-y)hatj+2hatk}-`
`{(2-x)hati+2hatj+2hatk}=xhati-yhatj`
`vec(AC)=vec(OC)-vec(OA)`
`={(2hati+2hatj+(2-z)hatk}-{(2-x)hati+2hatj+2hatk}`
`=xhati-zhatk`
`vec(AD)=vec(OD)-vec(OA)`
`=(hati+hatj+hatk)-{(2-x)hati+2hatj+2hatk}=(x-1)hati-hatj-hatk`
As these vectors are coplanar, `|(x, -y, 0),(x, 0, -z),(x-1, -1, -1)|=0`
On simplication we get `1/x+1/y+1/z=1`
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