If the lines $$\vec{r} = (\hat{i} - \hat{j} + \hat{k}) + \lambda(3\hat{j} - \hat{k})$$ and $$\vec{r} = (\alpha\hat{i} - \hat{j}) + \mu(2\hat{i} - 3\hat{k})$$ are co-planar, then the distance of the plane containing these two lines from the point $$(\alpha, 0, 0)$$ is

First, Line $$L_1$$ is described by $$\vec{r} = (\hat{i} - \hat{j} + \hat{k}) + \lambda(3\hat{j} - \hat{k})$$, so that point $$A = (1, -1, 1)$$ and direction $$\vec{d_1} = (0, 3, -1)$$. Next, Line $$L_2$$ is given by $$\vec{r} = (\alpha\hat{i} - \hat{j}) + \mu(2\hat{i} - 3\hat{k})$$, which yields point $$B = (\alpha, -1, 0)$$ and direction $$\vec{d_2} = (2, 0, -3)$$.

Since the lines are coplanar if $$(\vec{AB}) \cdot (\vec{d_1} \times \vec{d_2}) = 0$$, we compute $$\vec{AB} = (\alpha - 1, 0, -1)$$ and then evaluate $$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 3 & -1 \\ 2 & 0 & -3 \end{vmatrix}$$ which simplifies to $$\hat{i}(-9) - \hat{j}(2) + \hat{k}(-6) = (-9, -2, -6)$$.

Substituting these into the dot product gives $$\vec{AB} \cdot (\vec{d_1} \times \vec{d_2}) = (\alpha - 1)(-9) + 0(-2) + (-1)(-6) = -9(\alpha - 1) + 6 = -9\alpha + 9 + 6 = 15 - 9\alpha$$. Setting this to zero yields $$\alpha = \frac{15}{9} = \frac{5}{3}$$.

Now, the plane containing point $$A = (1, -1, 1)$$ has normal vector $$\vec{n} = (-9, -2, -6)$$, which is equivalent to $$(9, 2, 6)$$. Therefore the equation of the plane is $$9(x-1) + 2(y+1) + 6(z-1) = 0$$, leading to $$9x + 2y + 6z - 9 + 2 - 6 = 0$$ and hence $$9x + 2y + 6z - 13 = 0$$.

Next, the distance from the point $$(\alpha, 0, 0) = \left(\frac{5}{3}, 0, 0\right)$$ to this plane is given by $$d = \frac{\left|9 \cdot \frac{5}{3} + 2 \cdot 0 + 6 \cdot 0 - 13\right|}{\sqrt{81 + 4 + 36}} = \frac{|15 - 13|}{\sqrt{121}} = \frac{2}{11}$$.

Therefore the correct answer is Option B: $$\dfrac{2}{11}$$.