Let $$\tan\alpha$$, $$\tan\beta$$ and $$\tan\gamma$$; $$\alpha, \beta, \gamma \neq \frac{(2n-1)\pi}{2}$$, $$n \in N$$ be the slopes of the three line segments $$OA$$, $$OB$$ and $$OC$$, respectively, where $$O$$ is origin. If circumcentre of $$\triangle ABC$$ coincides with origin and its orthocentre lies on $$y$$-axis, then the value of $$\left(\frac{\cos 3\alpha + \cos 3\beta + \cos 3\gamma}{\cos\alpha \cdot \cos\beta \cdot \cos\gamma}\right)^2$$ is equal to ________.

Since the circumcentre of $$\triangle ABC$$ is at the origin $$O$$, we have $$|OA| = |OB| = |OC| = R$$ (circumradius). The lines $$OA$$, $$OB$$, $$OC$$ have slopes $$\tan\alpha$$, $$\tan\beta$$, $$\tan\gamma$$ respectively, so we can write $$A = R(\cos\alpha, \sin\alpha)$$, $$B = R(\cos\beta, \sin\beta)$$, $$C = R(\cos\gamma, \sin\gamma)$$.

The orthocentre of a triangle inscribed in a circle of radius $$R$$ centred at the origin is $$H = A + B + C = R(\cos\alpha + \cos\beta + \cos\gamma,\; \sin\alpha + \sin\beta + \sin\gamma)$$.

Since the orthocentre lies on the $$y$$-axis, its $$x$$-coordinate is zero: $$\cos\alpha + \cos\beta + \cos\gamma = 0$$.

Now we evaluate the given expression. Using the identity $$\cos 3\theta = 4\cos^3\theta - 3\cos\theta$$, we get $$\cos 3\alpha + \cos 3\beta + \cos 3\gamma = 4(\cos^3\alpha + \cos^3\beta + \cos^3\gamma) - 3(\cos\alpha + \cos\beta + \cos\gamma)$$.

Since $$\cos\alpha + \cos\beta + \cos\gamma = 0$$, the second term vanishes. For the first term, we use the identity: if $$p + q + r = 0$$, then $$p^3 + q^3 + r^3 = 3pqr$$. With $$p = \cos\alpha$$, $$q = \cos\beta$$, $$r = \cos\gamma$$, we get $$\cos^3\alpha + \cos^3\beta + \cos^3\gamma = 3\cos\alpha\cos\beta\cos\gamma$$.

Therefore $$\cos 3\alpha + \cos 3\beta + \cos 3\gamma = 4 \cdot 3\cos\alpha\cos\beta\cos\gamma = 12\cos\alpha\cos\beta\cos\gamma$$.

The required expression is $$\frac{\cos 3\alpha + \cos 3\beta + \cos 3\gamma}{\cos\alpha \cdot \cos\beta \cdot \cos\gamma} = \frac{12\cos\alpha\cos\beta\cos\gamma}{\cos\alpha\cos\beta\cos\gamma} = 12$$.

Therefore $$\left(\frac{\cos 3\alpha + \cos 3\beta + \cos 3\gamma}{\cos\alpha \cdot \cos\beta \cdot \cos\gamma}\right)^2 = 12^2 = 144$$.