If the shortest distance between the line joining the points $$(1, 2, 3)$$ and $$(2, 3, 4)$$, and the line $$\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-2}{0}$$ is $$\alpha$$, then $$28\alpha^2$$ is equal to _____.

Find $$28\alpha^2$$ where $$\alpha$$ is the shortest distance between the line joining $$(1,2,3)$$ and $$(2,3,4)$$, and the line $$\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-2}{0}$$.

Line $$L_1$$ passes through $$(1,2,3)$$ with direction $$(2-1,3-2,4-3)=(1,1,1)$$, so its equation is $$L_1:\frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{1}$$.

The second line is $$L_2:\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-2}{0}$$, which passes through $$(1,-1,2)$$ with direction $$(2,-1,0)$$.

To find the shortest distance we compute the cross product of the direction vectors $$\vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\1&1&1\\2&-1&0\end{vmatrix}=\hat i(0+1)-\hat j(0-2)+\hat k(-1-2)=(1,2,-3)$$ and the vector connecting points on the two lines is $$(\vec a_2-\vec a_1)=(1-1,-1-2,2-3)=(0,-3,-1)\,. $$

Hence the shortest distance is$$\alpha=\frac{|(\vec a_2-\vec a_1)\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}=\frac{|0\cdot1+(-3)\cdot2+(-1)\cdot(-3)|}{\sqrt{1+4+9}}=\frac{3}{\sqrt{14}}\,.$$

It follows that $$\alpha^2=\frac{9}{14}$$ and therefore $$28\alpha^2=28\times\frac{9}{14}=18\,. $$ The answer is $$\boxed{18}$$.