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

If `P(alpha,beta,lambda)` is a vertex of an equilateral triangle PQR w

1.	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 planeA. `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» Correct Answer - D 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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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 planeA. `x+y+z+6=0`B. `2x+4y+3z+20=0`C. `x-y+z+12=0`D. `x+y+z+3sqrt(2)=0`

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