Let the point $$(-1, \alpha, \beta)$$ lie on the line of the shortest distance between the lines $$\frac{x+2}{-3} = \frac{y-2}{4} = \frac{z-5}{2}$$ and $$\frac{x+2}{-1} = \frac{y+6}{2} = \frac{z-1}{0}$$. Then $$(\alpha - \beta)^2$$ is equal to __________

The two given skew lines are

$$\dfrac{x+2}{-3}=\dfrac{y-2}{4}=\dfrac{z-5}{2}$$
and
$$\dfrac{x+2}{-1}=\dfrac{y+6}{2}=\dfrac{z-1}{0}\;.$$

Write each line in vector form:

$$\mathbf{r}_1=(-2,\,2,\,5)+t\,(-3,\,4,\,2)$$
$$\mathbf{r}_2=(-2,\,-6,\,1)+s\,(-1,\,2,\,0)\;.$$

The direction vectors are $$\mathbf{a}=(-3,\,4,\,2)$$ and $$\mathbf{b}=(-1,\,2,\,0)$$.
The line of shortest distance is perpendicular to both lines, so its direction vector is

$$\mathbf{n}=\mathbf{a}\times\mathbf{b}\;.$$

Compute the cross-product:

$$\mathbf{n}=(-4,\,-2,\,-2)\;.$$

For convenience, use the proportional direction $$\mathbf{n}=(2,\,1,\,1)$$.

Let $$P=(-2-3t,\;2+4t,\;5+2t)$$ be a point on the first line and
$$Q=(-2-s,\;-6+2s,\;1)$$ be a point on the second line.

The vector $$\overrightarrow{PQ}=Q-P$$ must be parallel to $$\mathbf{n}$$:

$$\overrightarrow{PQ}=(3t-s,\,-8+2s-4t,\,-4-2t)=k\,(2,\,1,\,1)\;.$$

Equating components gives the system

$$3t-s=2k\;,$$
$$-8+2s-4t=k\;,$$
$$-4-2t=k\;.$$

Set the last two equal:

$$-8+2s-4t=-4-2t\;\Longrightarrow\;s-t-2=0\;\Longrightarrow\;s=t+2\;.$$

From $$k=-4-2t$$ (third equation) and the first equation, substitute $$s=t+2$$:

$$3t-(t+2)=2(-4-2t)\;.$$

Simplify:

$$2t-2=-8-4t\;\Longrightarrow\;6t=-6\;\Longrightarrow\;t=-1\;.$$

Hence $$s=t+2=1$$ and $$k=-4-2t=-2\;.$$

Thus the point on the first line is

$$P=(-2-3(-1),\;2+4(-1),\;5+2(-1))=(1,\,-2,\,3)\;,$$

and the required common perpendicular (shortest-distance) line is

$$\mathbf{r}=(1,\,-2,\,3)+\lambda\,(2,\,1,\,1)\;.$$

The point $$(-1,\alpha,\beta)$$ lies on this line, so

$$1+2\lambda=-1,\quad -2+\lambda=\alpha,\quad 3+\lambda=\beta\;.$$

From $$1+2\lambda=-1$$ we get $$\lambda=-1$$. Substituting:

$$\alpha=-2+(-1)=-3,\qquad \beta=3+(-1)=2\;.$$

Therefore

$$(\alpha-\beta)^2=(-3-2)^2=(-5)^2=25\;.$$

Hence, $$(\alpha-\beta)^2 = 25.$$