If $$\sin\left(\frac{\pi}{18}\right) \sin\left(\frac{5\pi}{18}\right) \sin\left(\frac{7\pi}{18}\right) = K$$, then the value of $$\sin\left(\frac{10K\pi}{3}\right)$$ is :

We note that $$\frac{\pi}{18} = 10°$$, $$\frac{5\pi}{18} = 50°$$, and $$\frac{7\pi}{18} = 70°$$. So we need $$K = \sin 10° \sin 50° \sin 70°$$.

Using the identity $$\sin\theta \sin(60° - \theta) \sin(60° + \theta) = \frac{1}{4}\sin 3\theta$$, with $$\theta = 10°$$:

$$\sin 10° \sin 50° \sin 70° = \frac{1}{4}\sin 30° = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}$$

So $$K = \frac{1}{8}$$.

Now we compute:

$$\sin\left(\frac{10K\pi}{3}\right) = \sin\left(\frac{10 \times \frac{1}{8} \times \pi}{3}\right) = \sin\left(\frac{10\pi}{24}\right) = \sin\left(\frac{5\pi}{12}\right) = \sin 75°$$

Using the compound angle formula:

$$\sin 75° = \sin(45° + 30°) = \sin 45° \cos 30° + \cos 45° \sin 30°$$

$$= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{\sqrt{3} + 1}{2\sqrt{2}}$$

Hence, the correct answer is Option 1.