Let √3^i+^j, ^i+√3^j and β^i+(1−β)^j respectively be the position vectors of the points A, B and C with respect to the origin O. If the distance of C from the bisector of the acute angle between OA and OB is 3√2, then the sum of all possible values of β is?

O=(0,0,0) → OA=√3^i+^j → OB=^i+√3^j → OC=β^i+(1−β)^j
Let the angle between OA and OB be θ. Then
cosθ = (OA.OB)/(|OA||OB|) = (√3+√3)/(2*2) = √3/2
θ = π/6
The equation of the bisector of the angle between OA and OB is given by
(x/√3 + y)/2 = (x+y√3)/2
√3x+y=x+√3y
(√3−1)x=(√3−1)y
Therefore, x=y
The distance of C from the bisector is given by
|β−(1−β)|/√2 = 3√2
|2β−1|/√2 = 3√2
|2β−1| = 6
2β−1=6 or 2β−1=−6
2β=7 or 2β=−5
β=7/2 or β=−5/2
Sum of all possible values of β = 7/2 + (-5/2) = 1