Let $$S=\left\{\sin ^2 2 \theta:\left(\sin ^4 \theta+\cos ^4 \theta\right) x^2+(\sin 2 \theta) x+\left(\sin ^6 \theta+\cos ^6 \theta\right)=0\right.$$ has real roots $\}$. If $\alpha$ and $\beta$ be the smallest and largest elements of the set $S$, respectively, then $3\left((\alpha-2)^2+(\beta-1)^2\right)$ equals __________.

<p>For real roots</p> <p>$$\begin{aligned} & D \geq 0 \\ & \sin ^2 2 \theta \geq 4\left(\sin ^4 \theta+\cos ^4 \theta\right)\left(\sin ^6 \theta+\cos ^6 \theta\right) \end{aligned}$$</p> <p>Put $\sin ^2 2 \theta=t$</p> <p>$$\begin{aligned} & \Rightarrow t \geq 4\left(1-\frac{t}{2}\right)\left(1-\frac{3 t}{4}\right) \\ & 2 t \geq(2-t)(4-3 t) \\ & 3 t^2-12 t+8 \leq 0 \\ & t^2-4 t+\frac{8}{3} \leq 0 \\ & (t-2)^2+\frac{8}{3}-4 \leq 0 \\ & (t-2)^2 \leq \frac{4}{3} \\ & -\frac{2}{\sqrt{3}} \leq t-2 \leq \frac{2}{\sqrt{3}} \\ & 2-\frac{2}{\sqrt{3}} \leq t \leq 2+\frac{2}{\sqrt{3}} \\ & \because t \in[0,1] \\ & \Rightarrow 2-\frac{2}{\sqrt{3}} \leq t \leq 1 \\ & \alpha=2-\frac{2}{\sqrt{3}}, \beta=1 \\ & \Rightarrow 3\left[(\alpha-2)^2+(\beta-1)^2\right]=4 \end{aligned}$$</p>