If the shortest distance between the lines $$\frac{x+\sqrt{6}}{2} = \frac{y-\sqrt{6}}{3} = \frac{z-\sqrt{6}}{4}$$ and $$\frac{x-\lambda}{3} = \frac{y-2\sqrt{6}}{4} = \frac{z+2\sqrt{6}}{5}$$ is 6, then square of sum of all possible values of $$\lambda$$ is

We need to find the square of the sum of all possible values of $$\lambda$$ such that the shortest distance between the two given lines is 6.

Line $$L_1$$: passes through $$(-\sqrt{6}, \sqrt{6}, \sqrt{6})$$ with direction $$(2, 3, 4)$$.

Line $$L_2$$: passes through $$(\lambda, 2\sqrt{6}, -2\sqrt{6})$$ with direction $$(3, 4, 5)$$.

$$\vec{d_1} \times \vec{d_2} = (2, 3, 4) \times (3, 4, 5)$$

$$= (15 - 16, 12 - 10, 8 - 9) = (-1, 2, -1)$$

$$|\vec{d_1} \times \vec{d_2}| = \sqrt{1 + 4 + 1} = \sqrt{6}$$

$$\vec{P_1P_2} = (\lambda + \sqrt{6}, 2\sqrt{6} - \sqrt{6}, -2\sqrt{6} - \sqrt{6}) = (\lambda + \sqrt{6}, \sqrt{6}, -3\sqrt{6})$$

$$d = \frac{|\vec{P_1P_2} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}$$

$$\vec{P_1P_2} \cdot (-1, 2, -1) = -(\lambda + \sqrt{6}) + 2\sqrt{6} + 3\sqrt{6} = -\lambda + 4\sqrt{6}$$

$$6 = \frac{|-\lambda + 4\sqrt{6}|}{\sqrt{6}} \implies |-\lambda + 4\sqrt{6}| = 6\sqrt{6}$$

$$-\lambda + 4\sqrt{6} = \pm 6\sqrt{6}$$

$$\lambda = 4\sqrt{6} - 6\sqrt{6} = -2\sqrt{6} \quad \text{or} \quad \lambda = 4\sqrt{6} + 6\sqrt{6} = 10\sqrt{6}$$

$$(\lambda_1 + \lambda_2)^2 = (-2\sqrt{6} + 10\sqrt{6})^2 = (8\sqrt{6})^2 = 384$$

The correct answer is $$384$$.