Let tan α ,   tan β and tan γ ;   α , β , γ ≠ ( 2 n - 1 )…

Given, the slopes of the line segments $OA, OB$ and $OC$ are respectively, $\tan \alpha , \tan \beta \& \tan \gamma ,$ then by parametric form the coordinates of the points $A, B \& C$ can be taken respectively as $\left(OA\cos \alpha , OA\sin \alpha \right), \left(OB\cos \beta , OB\sin \beta \right) \& \left(OC\cos \gamma , OC\sin \gamma \right).$

Given, the circumcentre of the $∆ABC$ 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 $\left(x_1, y_1\right), \left(x_2, y_2\right)$ and $\left(x_3, y_3\right)$ is $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right),$ thus the centroid of the $∆ABC$ is $\left(\frac{OA\left(\sum \cos \alpha \right)}{3}, \frac{OA\left(\sum \sin \alpha \right)}{3}\right).$

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.

$\Rightarrow \sum \cos \alpha =0$

$\Rightarrow \cos \alpha +\cos \beta +\cos \gamma =0$

We know that, if $a+b+c=0,$ then $a^3+b^3+c^3=3abc,$

$\Rightarrow {\cos }^3\alpha +{\cos }^3\beta +{\cos }^3\gamma =3·\cos \alpha ·\cos \beta ·\cos \gamma$

Now, using $\cos 3A=4{\cos }^{3}A-3\cos A,$ we have

${\left(\frac{\cos 3\alpha +\cos 3\beta +\cos 3\gamma }{\cos \alpha ·\cos \beta ·\cos \gamma }\right)}^2={\left(\frac{4\left({\cos }^3\alpha +{\cos }^3\beta +{\cos }^3\gamma \right)-3(\cos \alpha +\cos \beta +\cos \gamma )}{\cos \alpha ·\cos \beta ·\cos \gamma }\right)}^2$

$={\left(\frac{4\left(3·\cos \alpha ·\cos \beta ·\cos \gamma \right)-3\left(0\right)}{\cos \alpha ·\cos \beta ·\cos \gamma }\right)}^2$

$={\left(12\right)}^2=144.$