The sum of all values of $$\alpha$$, for which the sho1test distance between the lines
$$\frac{x+1}{\alpha}=\frac{y-2}{-1}=\frac{z-4}{-\alpha}$$ and $$\frac{x}{\alpha}=\frac{y-1}{2}=\frac{z-1}{2\alpha}$$ is $$\sqrt{2}$$, is

We need to find the sum of all values of $$\alpha$$ for which the shortest distance between the two lines is $$\sqrt{2}$$.

The first line is $$L_1: \frac{x+1}{\alpha} = \frac{y-2}{-1} = \frac{z-4}{-\alpha}$$, passing through $$(-1,2,4)$$ with direction vector $$(\alpha,-1,-\alpha)$$, and the second line is $$L_2: \frac{x}{\alpha} = \frac{y-1}{2} = \frac{z-1}{2\alpha}$$, passing through $$(0,1,1)$$ with direction vector $$(\alpha,2,2\alpha)$$.

$$\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & -1 & -\alpha \\ \alpha & 2 & 2\alpha \end{vmatrix}$$

$$= \hat{i}(-2\alpha + 2\alpha) - \hat{j}(2\alpha^2 + \alpha^2) + \hat{k}(2\alpha + \alpha)$$

$$= 0\hat{i} - 3\alpha^2\hat{j} + 3\alpha\hat{k}$$

$$= (0, -3\alpha^2, 3\alpha)$$

$$|\vec{d_1} \times \vec{d_2}| = \sqrt{0 + 9\alpha^4 + 9\alpha^2} = 3|\alpha|\sqrt{\alpha^2 + 1}$$

$$\vec{P_1P_2} = (0-(-1), 1-2, 1-4) = (1, -1, -3)$$

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

$$\vec{P_1P_2} \cdot (0, -3\alpha^2, 3\alpha) = 0 + 3\alpha^2 - 9\alpha = 3\alpha(\alpha - 3)$$

$$d = \frac{|3\alpha(\alpha - 3)|}{3|\alpha|\sqrt{\alpha^2 + 1}} = \frac{|\alpha - 3|}{\sqrt{\alpha^2 + 1}}$$

Setting the distance equal to $$\sqrt{2}$$ gives

$$\frac{|\alpha - 3|}{\sqrt{\alpha^2 + 1}} = \sqrt{2}$$

Squaring both sides gives

$$\frac{(\alpha - 3)^2}{\alpha^2 + 1} = 2$$

$$\alpha^2 - 6\alpha + 9 = 2\alpha^2 + 2$$

$$\alpha^2 + 6\alpha - 7 = 0$$

$$(\ \alpha + 7)(\alpha - 1) = 0$$

$$\alpha = -7$$ or $$\alpha = 1$$

The sum of all values is $$-7 + 1 = -6$$.

Therefore, the answer is Option 4: -6.