If the angle between the line $$2(x+1) = y = z + 4$$ and the plane $$2x - y + \sqrt{\lambda}z + 4 = 0$$ is $$\frac{\pi}{6}$$, then the value of $$\lambda$$ is:

The given line is $$2(x + 1) = y = z + 4$$. To find the direction ratios, express the line in symmetric form. Set $$2(x + 1) = y = z + 4 = k$$, where $$k$$ is a parameter. Then, $$2(x + 1) = k$$ gives $$x + 1 = \frac{k}{2}$$, so $$x = \frac{k}{2} - 1$$. Also, $$y = k$$ and $$z + 4 = k$$ gives $$z = k - 4$$. The direction ratios are the coefficients of $$k$$: $$\frac{1}{2}$$ for $$x$$, $$1$$ for $$y$$, and $$1$$ for $$z$$. To avoid fractions, multiply by 2, giving direction ratios $$(1, 2, 2)$$. Thus, the direction vector of the line is $$\vec{d} = (1, 2, 2)$$.

The plane equation is $$2x - y + \sqrt{\lambda} z + 4 = 0$$. The normal vector to the plane is $$\vec{n} = (2, -1, \sqrt{\lambda})$$.

The angle between the line and the plane is given as $$\frac{\pi}{6}$$. The formula for the sine of the angle $$\theta$$ between a line and a plane is $$\sin \theta = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}| |\vec{n}|}$$. Here, $$\theta = \frac{\pi}{6}$$, so $$\sin \frac{\pi}{6} = \frac{1}{2}$$.

Compute the dot product $$\vec{d} \cdot \vec{n}$$: $$$ \vec{d} \cdot \vec{n} = (1)(2) + (2)(-1) + (2)(\sqrt{\lambda}) = 2 - 2 + 2\sqrt{\lambda} = 2\sqrt{\lambda}. $$$ Since $$\lambda$$ must be non-negative (as it is under a square root), $$|\vec{d} \cdot \vec{n}| = |2\sqrt{\lambda}| = 2\sqrt{\lambda}$$.

Compute the magnitudes: $$$ |\vec{d}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3, $$$ $$$ |\vec{n}| = \sqrt{2^2 + (-1)^2 + (\sqrt{\lambda})^2} = \sqrt{4 + 1 + \lambda} = \sqrt{5 + \lambda}. $$$

Substitute into the formula: $$$ \sin \theta = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}| |\vec{n}|} \implies \frac{1}{2} = \frac{2\sqrt{\lambda}}{3 \sqrt{5 + \lambda}}. $$$

Solve for $$\lambda$$: $$$ \frac{1}{2} = \frac{2\sqrt{\lambda}}{3 \sqrt{5 + \lambda}}. $$$

Multiply both sides by $$3 \sqrt{5 + \lambda}$$: $$$ \frac{1}{2} \cdot 3 \sqrt{5 + \lambda} = 2\sqrt{\lambda} \implies \frac{3}{2} \sqrt{5 + \lambda} = 2\sqrt{\lambda}. $$$

Divide both sides by 2: $$$ \frac{3}{4} \sqrt{5 + \lambda} = \sqrt{\lambda}. $$$

Square both sides to eliminate the square roots: $$$ \left(\frac{3}{4}\right)^2 (5 + \lambda) = (\sqrt{\lambda})^2 \implies \frac{9}{16} (5 + \lambda) = \lambda. $$$

Multiply both sides by 16: $$$ 9(5 + \lambda) = 16\lambda \implies 45 + 9\lambda = 16\lambda. $$$

Rearrange terms: $$$ 45 = 16\lambda - 9\lambda \implies 45 = 7\lambda \implies \lambda = \frac{45}{7}. $$$

Verify the solution by substituting $$\lambda = \frac{45}{7}$$ back into the equation before squaring. Left side: $$\frac{3}{4} \sqrt{5 + \frac{45}{7}} = \frac{3}{4} \sqrt{\frac{35}{7} + \frac{45}{7}} = \frac{3}{4} \sqrt{\frac{80}{7}} = \frac{3}{4} \cdot \frac{\sqrt{80}}{\sqrt{7}} = \frac{3}{4} \cdot \frac{4\sqrt{5}}{\sqrt{7}} = \frac{3\sqrt{5}}{\sqrt{7}}$$. Right side: $$\sqrt{\frac{45}{7}} = \frac{\sqrt{45}}{\sqrt{7}} = \frac{3\sqrt{5}}{\sqrt{7}}$$. Both sides are equal, so the solution is valid.

Comparing with the options, $$\lambda = \frac{45}{7}$$ corresponds to Option A.

Hence, the correct answer is Option A.