Let tanα, tanβ and tanγ; α, β, γ ≠ (( 2n - 1 ) π)/(2), n ∈ N be the...

Given, the slopes of the line segments $OA, OB and OC$ are respectively, $tan\alpha, tan\beta \text{\text{ and } } tan\gamma,$ then by parametric form the coordinates of the points $A, B \text{\text{ and } }$ C can be taken respectively as \left(OAcos\alpha, OAsin\alpha\right), \left(OBcos\beta, OBsin\beta\right) \text{ and } \left(OCcos\gamma, OCsin\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 $OA = OB = OC .</p><p>The centroid of a triangle with vertices \left({x}_{1}, {y}_{1}\right),$\left({x}_{2}, {y}_{2}\right)$and \left({x}_{3}, {y}_{3}\right)$ is \left(\dfrac{{x}_{1} + {x}_{2} + {x}_{3}}{3}, \dfrac{{y}_{1} + {y}_{2} + {y}_{3}}{3}\right),$thus the centroid of the$∆ ABC$is \left(\dfrac{OA\left(∑ cos\alpha\right)}{3}, \dfrac{OA\left(∑ sin\alpha\right)}{3}\right) .</p><p>We know that$ for any triangle the circumcentre, orthocentre and centroid lie on a line.

Since, the orthocentre and circumcentre both lies ony$-axis, hence the centroid also lies on$y$-axis, hence the$x$-coordinate of the centroid is zero.</p><p>$⇒ ∑ cos\alpha = 0$</p><p>$⇒ cos\alpha + cos\beta + cos\gamma = 0$</p><p>We know that, if$a + b + c = 0,$then${a}^{3} + {b}^{3} + {c}^{3} = 3abc,$</p><p>$⇒ \cos^{3}\alpha + \cos^{3}\beta + \cos^{3}\gamma = 3 \cdot cos\alpha \cdot cos\beta \cdot cos\gamma$</p><p>Now, using$\cos 3A = 4\cos^{3}A - 3cosA,$we have</p><p>${\left(\dfrac{\cos 3\alpha + \cos 3\beta + \cos 3\gamma}{cos\alpha \cdot cos\beta \cdot cos\gamma}\right)}^{2} = {\left(\dfrac{4\left(\cos^{3}\alpha + \cos^{3}\beta + \cos^{3}\gamma\right) - 3 (\text{ cos }\alpha +\text{ cos }\beta +\text{ cos }\gamma )}{cos\alpha \cdot cos\beta \cdot cos\gamma}\right)}^{2}$</p><p>$= {\left(\dfrac{4$\left(3 \cdot cos\alpha \cdot cos\beta \cdot cos\gamma\right)$ - 3\left(0\right)}{cos\alpha \cdot cos\beta \cdot cos\gamma}\right)}^{2}

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