Let the shortest distance between the lines L: $$\dfrac{x-5}{-2} = \dfrac{y-\lambda}{0} = \dfrac{z+\lambda}{1}$$, $$\lambda \ge 0$$ and L$$_1$$: $$x+1 = y-1 = 4-z$$ be $$2\sqrt{6}$$. If $$(\alpha, \beta, \gamma)$$ lies on L, then which of the following is NOT possible?

Lines: $$L: \frac{x-5}{-2} = \frac{y-\lambda}{0} = \frac{z+\lambda}{1}$$ and $$L_1: \frac{x+1}{1} = \frac{y-1}{1} = \frac{z-4}{-1}$$.

To find the shortest distance between these lines, consider a point on $$L$$ as $$(5,\lambda,-\lambda)$$ with direction vector $$\vec{d_1} = (-2, 0, 1)$$, and a point on $$L_1$$ as $$(-1,1,4)$$ with direction vector $$\vec{d_2} = (1, 1, -1)$$.

The cross product of the direction vectors is $$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 0 & 1 \\ 1 & 1 & -1 \end{vmatrix} = \hat{i}(0-1) - \hat{j}(2-1) + \hat{k}(-2-0) = (-1, -1, -2),$$ and its magnitude is $$|\vec{d_1} \times \vec{d_2}| = \sqrt{1+1+4} = \sqrt{6}.$$

The vector connecting the two chosen points is $$\vec{PQ} = (5-(-1),\,\lambda-1,\,-\lambda-4) = (6,\;\lambda-1,\;-\lambda-4).$$ Hence the shortest distance is given by $$\frac{|\vec{PQ}\cdot(\vec{d_1}\times\vec{d_2})|}{|\vec{d_1}\times\vec{d_2}|} = \frac{|6(-1) + (\lambda-1)(-1) + (-\lambda-4)(-2)|}{\sqrt{6}} = \frac{|-6 - \lambda + 1 + 2\lambda + 8|}{\sqrt{6}} = \frac{|\lambda + 3|}{\sqrt{6}}.$$

Given that the shortest distance equals $$2\sqrt{6}$$, we have $$\frac{|\lambda + 3|}{\sqrt{6}} = 2\sqrt{6} \;\Rightarrow\; |\lambda + 3| = 12.$$ Since $$\lambda \ge 0$$, it follows that $$\lambda + 3 = 12$$ so $$\lambda = 9$$ (the solution $$\lambda = -15$$ is rejected).

Substituting $$\lambda = 9$$ into the parametric form of $$L$$ gives $$\frac{x-5}{-2} = \frac{y-9}{0} = \frac{z+9}{1} = t,$$ so points on $$L$$ can be written as $$(\alpha,\beta,\gamma) = (5-2t,\;9,\;-9+t).$$ Since $$\beta=9$$ and $$\alpha=5-2t,\;\gamma=-9+t$$, one finds $$t=\gamma+9$$ and thus $$\alpha = 5 - 2(\gamma+9) = -13 - 2\gamma,$$ yielding the constant relation $$\alpha + 2\gamma = -13.$$

Checking the options shows: (1) $$\alpha + 2\gamma = 24$$ is not possible since $$-13\neq24$$. (2) $$2\alpha + \gamma = 2(-13-2\gamma) + \gamma = -26 - 3\gamma$$, and setting this equal to 7 gives $$\gamma = -11$$, which is possible. (3) $$2\alpha - \gamma = 2(-13-2\gamma) - \gamma = -26 - 5\gamma$$, and setting this to 9 yields $$\gamma = -7$$, possible. (4) $$\alpha - 2\gamma = -13 - 4\gamma$$, and setting this to 19 gives $$\gamma = -8$$, possible.

The correct answer is Option (1): $$\boxed{\alpha + 2\gamma = 24}$$ is not possible.