Solution:
\( S = \left\{ \sin^{2} 2\theta : \; \left( \sin^{4}\theta + \cos^{4}\theta \right)x^{2} + \left( \sin 2\theta \right)x + \left( \sin^{6}\theta + \cos^{6}\theta \right) = 0 \text{ has real roots} \right\} \)
\( \sin^{4}\theta + \cos^{4}\theta = (\sin^{2}\theta + \cos^{2}\theta)^{2} - 2\sin^{2}\theta\cos^{2}\theta = 1 - \dfrac{1}{2}\sin^{2} 2\theta \)
\( \sin^{6}\theta + \cos^{6}\theta = (\sin^{2}\theta + \cos^{2}\theta)^{3} - 3\sin^{2}\theta\cos^{2}\theta (\sin^2\theta+cos^2\theta) = 1 - \dfrac{3}{4}\sin^{2} 2\theta \)
Let \( \sin 2\theta = t \)
\( \therefore \; \left(1 - \dfrac{{1}}{2}t^2\right)x^{2} + t x + \left(1 - \dfrac{3}{4}t^{2}\right) = 0 \)
For real roots \(\Rightarrow D \ge 0 \Rightarrow t^{2} - 4\left(1 - \dfrac{t^{2}}{2}\right)\left(1 - \dfrac{3}{4}t^{2}\right) \ge 0 \)
\(t^2 - (2 - t^2)\left(2 - \frac{3t^2}{2}\right) \ge 0\)
\(t^2 - 4 + 3t^2 + 2t^2 - \frac{3t^4}{2} \ge 0\)
\( \Rightarrow- \dfrac{3t^{4}}{2} + 6t^{2} - 4 \ge 0\) \( \Rightarrow 3t^{4} - 12t^{2} + 8 \le 0 \)
\( \Rightarrow (t^{2} - \alpha)(t^{2} - \beta) \le 0 \)
\( t^{2} = \dfrac{12 \pm \sqrt{144 - 4 \times 3 \times 8}}{2\times 3} = \dfrac{12 \pm \sqrt{48}}{6} = 2 \pm \dfrac{\sqrt{48}}{6} \)
\( 2 - \dfrac{\sqrt{48}}{6} \le \sin^{2} 2\theta \le 2 + \dfrac{\sqrt{48}}{6} \)
But \( \sin^{2} 2\theta \le 1 \),\( \therefore \; 2 - \dfrac{\sqrt{48}}{6} \le \sin^{2} 2\theta \le 1 \Rightarrow \alpha = 2 - \dfrac{\sqrt{48}}{6} \) & \( \beta = 1 \)
\( 3[(\alpha - 2)^{2} + (\beta - 1)^{2}] = 3\!\left\{\!\left(-\dfrac{\sqrt{48}}{6}\right)^{2} + 0\!\right\} = 3 \times \dfrac{48}{36} = 4 \)

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