Given, the slopes of the line segments OA,OB and OC are respectively, \mathrm{tan}\alpha ,\mathrm{tan}\beta &\mathrm{tan}\gamma , then by parametric form the coordinates of the points A,B&C can be taken respectively as (OA\mathrm{cos}\alpha ,OA\mathrm{sin}\alpha ),(OB\mathrm{cos}\beta ,OB\mathrm{sin}\beta )&(OC\mathrm{cos}\gamma ,OC\mathrm{sin}\gamma ).
Given, the circumcentre of the ΔABC is at origin and we know that the circumcentre is equidistant from the vertices of the triangle, hence OA=OB=OC.
The centroid of a triangle with vertices (x1,y1),(x2,y2) and (x3,y3) is (3x1+x2+x3,3y1+y2+y3), thus the centroid of the ΔABC is (3OA(∑cosα),3OA(∑sinα)).
We know that for any triangle the circumcentre, orthocentre and centroid lie on a line.
Since, the orthocentre and circumcentre both lies on y-axis, hence the centroid also lies on y-axis, hence the x-coordinate of the centroid is zero.
⇒∑cosα=0
⇒cosα+cosβ+cosγ=0
We know that, if a+b+c=0, then a3+b3+c3=3abc,
⇒cos3α+cos3β+cos3γ=3⋅cosα⋅cosβ⋅cosγ
Now, using cos3A=4cos3A−3cosA, we have
(cosα⋅cosβ⋅cosγcos3α+cos3β+cos3γ)2=(cosα⋅cosβ⋅cosγ4(cos3α+cos3β+cos3γ)−3(cosα+cosβ+cosγ))2
=(cosα⋅cosβ⋅cosγ4(3⋅cosα⋅cosβ⋅cosγ)−3(0))2
=(12)2=144.