Let the angle $$\theta, 0 < \theta < \frac{\pi}{2}$$ between two unit vectors $$\hat{a}$$ and $$\hat{b}$$ be $$\sin^{-1}\left(\frac{\sqrt{65}}{9}\right)$$. If the vector $$\vec{c} = 3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})$$, then the value of $$9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b})$$ is

Given $$\sin\theta = \frac{\sqrt{65}}{9}$$ where $$0 \lt \theta \lt \frac{\pi}{2}$$, so $$\cos\theta = \sqrt{1 - \frac{65}{81}} = \sqrt{\frac{16}{81}} = \frac{4}{9}$$.

The vector $$\vec{c} = 3\hat{a} + 6\hat{b} + 9(\hat{a} \times \hat{b})$$.

Computing $$\vec{c} \cdot \hat{a}$$: Since $$\hat{a} \cdot \hat{a} = 1$$, $$\hat{b} \cdot \hat{a} = \cos\theta = \frac{4}{9}$$, and $$(\hat{a} \times \hat{b}) \cdot \hat{a} = 0$$:

$$\vec{c} \cdot \hat{a} = 3(1) + 6\left(\frac{4}{9}\right) + 0 = 3 + \frac{8}{3} = \frac{17}{3}$$

Computing $$\vec{c} \cdot \hat{b}$$: Since $$\hat{a} \cdot \hat{b} = \cos\theta = \frac{4}{9}$$, $$\hat{b} \cdot \hat{b} = 1$$, and $$(\hat{a} \times \hat{b}) \cdot \hat{b} = 0$$:

$$\vec{c} \cdot \hat{b} = 3\left(\frac{4}{9}\right) + 6(1) + 0 = \frac{4}{3} + 6 = \frac{22}{3}$$

Therefore: $$9(\vec{c} \cdot \hat{a}) - 3(\vec{c} \cdot \hat{b}) = 9 \times \frac{17}{3} - 3 \times \frac{22}{3} = 51 - 22 = 29$$

Hence, the correct answer is Option C.