1. The problem asks to calculate $$\frac{9}{\sin^4 \alpha + \cos^4 \alpha}$$ given that $$\cot \alpha = \sqrt{3}$$. 2. Recall that $$\cot \alpha = \frac{\cos \alpha}{\sin \alpha}$$, so $$\frac{\cos \alpha}{\sin \alpha} = \sqrt{3}$$. 3. From this, we can express $$\cos \alpha = \sqrt{3} \sin \alpha$$. 4. Using the Pythagorean identity $$\sin^2 \alpha + \cos^2 \alpha = 1$$, substitute $$\cos \alpha$$: $$\sin^2 \alpha + (\sqrt{3} \sin \alpha)^2 = 1$$ $$\sin^2 \alpha + 3 \sin^2 \alpha = 1$$ $$4 \sin^2 \alpha = 1$$ $$\sin^2 \alpha = \frac{1}{4}$$ 5. Then $$\sin \alpha = \frac{1}{2}$$ (taking positive root as usual for angles in first quadrant). 6. Calculate $$\cos^2 \alpha = 1 - \sin^2 \alpha = 1 - \frac{1}{4} = \frac{3}{4}$$. 7. Now compute $$\sin^4 \alpha + \cos^4 \alpha = (\sin^2 \alpha)^2 + (\cos^2 \alpha)^2 = \left(\frac{1}{4}\right)^2 + \left(\frac{3}{4}\right)^2 = \frac{1}{16} + \frac{9}{16} = \frac{10}{16} = \frac{5}{8}$$. 8. Finally, calculate the expression: $$\frac{9}{\sin^4 \alpha + \cos^4 \alpha} = \frac{9}{\frac{5}{8}} = 9 \times \frac{8}{5} = \frac{72}{5} = 14.4$$. Answer: 14.4 (Option E).