Let $$\lambda_1, \lambda_2$$ be the values of $$\lambda$$ for which the points $$\left(\dfrac{5}{2}, 1, \lambda\right)$$ and $$(-2, 0, 1)$$ are at equal distance from the plane $$2x + 3y - 6z + 7$$. If $$\lambda_1 > \lambda_2$$ then the distance of the point $$(\lambda_1 - \lambda_2, \lambda_2, \lambda_1)$$ from the line $$\dfrac{x-5}{1} = \dfrac{y-1}{2} = \dfrac{z+7}{2}$$ is ______.

We need to find the values of $$\lambda$$ for which the points $$\left(\dfrac{5}{2}, 1, \lambda\right)$$ and $$(-2, 0, 1)$$ are equidistant from the plane $$2x + 3y - 6z + 7 = 0$$.

Distance = $$\dfrac{|2(-2) + 3(0) - 6(1) + 7|}{\sqrt{4 + 9 + 36}} = \dfrac{|-4 - 6 + 7|}{7} = \dfrac{|-3|}{7} = \dfrac{3}{7}$$

Distance = $$\dfrac{|2 \cdot \dfrac{5}{2} + 3(1) - 6\lambda + 7|}{7} = \dfrac{|5 + 3 - 6\lambda + 7|}{7} = \dfrac{|15 - 6\lambda|}{7}$$

$$|15 - 6\lambda| = 3$$

Case 1: $$15 - 6\lambda = 3 \Rightarrow \lambda = 2$$

Case 2: $$15 - 6\lambda = -3 \Rightarrow \lambda = 3$$

So $$\lambda_1 = 3$$ and $$\lambda_2 = 2$$ (since $$\lambda_1 > \lambda_2$$).

Find the point $$P = (\lambda_1 - \lambda_2, \lambda_2, \lambda_1) = (1, 2, 3)$$.

Point on line: $$A = (5, 1, -7)$$, direction vector: $$\vec{d} = (1, 2, 2)$$.

$$\vec{AP} = (1-5, 2-1, 3-(-7)) = (-4, 1, 10)$$

$$\vec{AP} \times \vec{d} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -4 & 1 & 10 \\ 1 & 2 & 2 \end{vmatrix}$$

$$= \vec{i}(2-20) - \vec{j}(-8-10) + \vec{k}(-8-1) = (-18, 18, -9)$$

$$|\vec{AP} \times \vec{d}| = \sqrt{324 + 324 + 81} = \sqrt{729} = 27$$

$$|\vec{d}| = \sqrt{1 + 4 + 4} = 3$$

Distance = $$\dfrac{27}{3} = 9$$

The answer is $$\boxed{9}$$.