Let $$ P(\alpha,\beta, \gamma)$$ be the point on the line $$\frac{x-1}{2}=\frac{y+1}{-3}=z$$ at a distance $$4\sqrt{14}$$ from the point (1, -1, 0) and nearer to the origin. Then the shortest di stance, between the Lines $$\frac{x-\alpha}{1}=\frac{y-\beta}{2}=\frac{z-\gamma}{3}$$ and $$\frac{x+5}{2}= \frac{y-10}{1}=\frac{z-3}{1}$$, is equal to

The point $$P(\alpha, \beta, \gamma)$$ lies on the line $$\frac{x-1}{2} = \frac{y+1}{-3} = z$$. Parametrize the line by setting $$\frac{x-1}{2} = \frac{y+1}{-3} = z = \lambda$$. Then:

$$x = 1 + 2\lambda, \quad y = -1 - 3\lambda, \quad z = \lambda$$

So, any point on the line is $$(1 + 2\lambda, -1 - 3\lambda, \lambda)$$. The distance from $$(1, -1, 0)$$ to this point is given as $$4\sqrt{14}$$. The distance formula gives:

$$\sqrt{(1 + 2\lambda - 1)^2 + (-1 - 3\lambda - (-1))^2 + (\lambda - 0)^2} = \sqrt{(2\lambda)^2 + (-3\lambda)^2 + \lambda^2} = \sqrt{4\lambda^2 + 9\lambda^2 + \lambda^2} = \sqrt{14\lambda^2} = \sqrt{14} |\lambda|$$

Set this equal to $$4\sqrt{14}$$:

$$\sqrt{14} |\lambda| = 4\sqrt{14} \implies |\lambda| = 4 \implies \lambda = 4 \quad \text{or} \quad \lambda = -4$$

The corresponding points are:

For $$\lambda = 4$$: $$(1 + 2(4), -1 - 3(4), 4) = (9, -13, 4)$$

For $$\lambda = -4$$: $$(1 + 2(-4), -1 - 3(-4), -4) = (-7, 11, -4)$$

Now, find which point is nearer to the origin $$(0, 0, 0)$$:

Distance to $$(9, -13, 4)$$: $$\sqrt{9^2 + (-13)^2 + 4^2} = \sqrt{81 + 169 + 16} = \sqrt{266}$$

Distance to $$(-7, 11, -4)$$: $$\sqrt{(-7)^2 + 11^2 + (-4)^2} = \sqrt{49 + 121 + 16} = \sqrt{186}$$

Since $$\sqrt{186} < \sqrt{266}$$, the point $$(-7, 11, -4)$$ is nearer to the origin. Thus, $$P(\alpha, \beta, \gamma) = (-7, 11, -4)$$.

Now, find the shortest distance between the lines:

Line 1: $$\frac{x - (-7)}{1} = \frac{y - 11}{2} = \frac{z - (-4)}{3} \implies \frac{x + 7}{1} = \frac{y - 11}{2} = \frac{z + 4}{3}$$

Line 2: $$\frac{x + 5}{2} = \frac{y - 10}{1} = \frac{z - 3}{1}$$

The direction vectors are:

$$\vec{d_1} = (1, 2, 3)$$ for Line 1

$$\vec{d_2} = (2, 1, 1)$$ for Line 2

A point on Line 1: $$A(-7, 11, -4)$$

A point on Line 2: Set $$\frac{x + 5}{2} = \frac{y - 10}{1} = \frac{z - 3}{1} = 0$$, so $$B(-5, 10, 3)$$

The vector $$\overrightarrow{AB}$$ is:

$$\overrightarrow{AB} = (-5 - (-7), 10 - 11, 3 - (-4)) = (2, -1, 7)$$

The shortest distance $$d$$ between two skew lines is given by:

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

First, compute $$\vec{d_1} \times \vec{d_2}$$:

$$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{vmatrix} = \hat{i}(2 \cdot 1 - 3 \cdot 1) - \hat{j}(1 \cdot 1 - 3 \cdot 2) + \hat{k}(1 \cdot 1 - 2 \cdot 2) = \hat{i}(2 - 3) - \hat{j}(1 - 6) + \hat{k}(1 - 4) = -\hat{i} + 5\hat{j} - 3\hat{k}$$

So, $$\vec{d_1} \times \vec{d_2} = (-1, 5, -3)$$

Magnitude: $$|\vec{d_1} \times \vec{d_2}| = \sqrt{(-1)^2 + 5^2 + (-3)^2} = \sqrt{1 + 25 + 9} = \sqrt{35}$$

Now, dot product:

$$\overrightarrow{AB} \cdot (\vec{d_1} \times \vec{d_2}) = (2)(-1) + (-1)(5) + (7)(-3) = -2 - 5 - 21 = -28$$

Absolute value: $$|-28| = 28$$

Thus,

$$d = \frac{28}{\sqrt{35}} = \frac{28}{\sqrt{35}} \cdot \frac{\sqrt{35}}{\sqrt{35}} = \frac{28\sqrt{35}}{35} = \frac{4\sqrt{35}}{5}$$

Simplify:

$$\frac{4\sqrt{35}}{5} = 4 \cdot \frac{\sqrt{35}}{5} = 4 \cdot \sqrt{\frac{35}{25}} = 4 \sqrt{\frac{7}{5}}$$

Therefore, the shortest distance is $$4\sqrt{\frac{7}{5}}$$, which corresponds to option B.