If the shortest distance between the lines $$\vec{r_1} = \alpha\hat{i} + 2\hat{j} + 2\hat{k} + \lambda(\hat{i} - 2\hat{j} + 2\hat{k})$$, $$\lambda \in R$$, $$\alpha > 0$$ and $$\vec{r_2} = -4\hat{i} - \hat{k} + \mu(3\hat{i} - 2\hat{j} - 2\hat{k})$$, $$\mu \in R$$ is 9, then $$\alpha$$ is equal to ___.

Line 1 passes through $$(\alpha, 2, 2)$$ with direction $$\vec{d_1} = \hat{i} - 2\hat{j} + 2\hat{k}$$. Line 2 passes through $$(-4, 0, -1)$$ with direction $$\vec{d_2} = 3\hat{i} - 2\hat{j} - 2\hat{k}$$.

We compute $$\vec{d_1} \times \vec{d_2}$$:

$$\vec{d_1} \times \vec{d_2} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-2&2\\3&-2&-2\end{vmatrix} = \hat{i}(4+4) - \hat{j}(-2-6) + \hat{k}(-2+6) = 8\hat{i} + 8\hat{j} + 4\hat{k}$$

$$|\vec{d_1} \times \vec{d_2}| = \sqrt{64 + 64 + 16} = \sqrt{144} = 12$$

The vector connecting the two base points is $$\vec{w} = (-4-\alpha, -2, -3)$$. The shortest distance formula gives:

$$\text{SD} = \frac{|\vec{w} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|} = \frac{|(-4-\alpha)(8) + (-2)(8) + (-3)(4)|}{12} = \frac{|-32 - 8\alpha - 16 - 12|}{12} = \frac{|60 + 8\alpha|}{12}$$

Setting the shortest distance equal to 9:

$$\frac{|60 + 8\alpha|}{12} = 9 \implies |60 + 8\alpha| = 108$$

This gives $$60 + 8\alpha = 108 \implies \alpha = 6$$, or $$60 + 8\alpha = -108 \implies \alpha = -21$$. Since $$\alpha > 0$$, we have $$\alpha = 6$$.