This implies, |AB|=√(α−β)2+(β−γ)2+(γ−α)2 And, |BC|=√(β−γ)2+(γ−α)2+(α−β)2 And |CA|=√(γ−α)2+(α−β)2+(β−γ)2 This implies, |AB|=|BC|=|CA|=√(α−β)2+(β−γ)2+(γ−α)2 So, the given position vectors are the vertices of equilateral triangle.