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If P(alpha,beta,lambda) 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

Answer»

`x+y+z+6=0`
`2x+4y+3z+20=0`
`x-y+z+12=0`
`x+y+z+3sqrt(2)=0`

Solution :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`


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