If `P(alpha,beta,gamma)` is a vertex of an equilateral triangle PQR where vertex Q and R are `(-1,0,1)` and `(1,0,-1)` respectively, then P can lie on the plane

Options

A. `x+y+z+6=0`
B. `2x+4y+3z+20=0`
C. `x-y+z+12=0`
D. `x+y+z+3sqrt(2)=0`

Answer

D

Explanation

We have `Q(-1,0,1)` and R(1,0,-1).
`therefore QR=2sqrt(2)`
Now, the mid point of QR is M(0,0,0).
Triangle PQR is equilateral.
`therefore PM=MQ xx tan60^(@) = sqrt(2)sqrt(3)=sqrt(6)`.
Thus, P can lie on the plane whose distance from point M is less than or equal to `sqrt(6)`.
Hence, P can lie on the plane `x+y+z+3sqrt(2)=0`