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The intersection of the spheres x^2+y...

The intersection of the spheres `x^2+y^2+z^2+7x-2y-z=13a n dx^2+y^2+z^2-3x+3y+4z=8` is the same as the intersection of one of the spheres and the plane a. `x-y-z=1` b. `x-2y-z=1` c. `x-y-2z=1` d. `2x-y-z=1`

A

`x-y-z=1`

B

`x-2y-z=1`

C

`x-y-2z=1`

D

`2x-y-z=1`

Text Solution

Verified by Experts

The given spheres are
`x^(2)+y^(2)+7x-2y-z-13=0" "(i)`
and `x^(2)+y^(2)+z^(2)-3x+3y+2-8=0" "(ii)`
Subtracting (ii) from (i), we get
`10x-5y-5z-5=0`
or `2x-y-z=1`
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