The equation of the plane passing through the points (1,1,1) and (1,-1,-1) and perpendicular to plane `2x-y+z=0` is

Options

A. `x+y+z+1=0`
B. `x+y-z-1=0`
C. `x+y-z+1=0`
D. `x+y+z-1=0`

Answer

B

Explanation

Any plane passing through (1,1,1) is
`a(x-1)+b(y-1)+c(z-1)=0` . . . . (i)
Since, it is passing through (1,-1,-1), we get
`a.0+b(-2)+c(-2)=0`
`rArr" "0.a-2b-2c=0`
`rArr" "0.a+b+c=0` . . . (ii)
Eqs (i) is perpendicular to the plane 2x-y+z+5=0
So, `2a-b+c=0" "[becausea_(a)a_(2)+b_(a)b_(2)+c_(a)c_(2)=0]` . . . .(iii)
After solving Eqs. (ii) and (iii) by method of cross-multiplication, we get
`a=b=2lamda,c=-2lamda`
On substituting the value of a,b and c in Eq. (i) we get
x+y-z-1=0