If the shortest distance between the lines $$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$$ and $$\frac{x}{1} = \frac{y}{\alpha} = \frac{z-5}{1}$$ is $$\frac{5}{\sqrt{6}}$$, then the sum of all possible values of $$\alpha$$ is

The first line is given in symmetric form as  $$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=t$$
Point on the line: $$A(1,2,3)$$
Direction vector: $$\mathbf{a}=2\mathbf{i}+3\mathbf{j}+4\mathbf{k}$$.

The second line is  $$\frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1}=s$$
Point on the line: $$B(0,0,5)$$
Direction vector: $$\mathbf{b}=1\mathbf{i}+\alpha\mathbf{j}+1\mathbf{k}$$.

Vector joining the two given points is  $$\mathbf{AB}=B-A=(-1,-2,2).$$

For two skew lines with direction vectors $$\mathbf{a}$$ and $$\mathbf{b}$$ and a joining vector $$\mathbf{c}$$, the shortest distance $$d$$ is
$$d=\frac{\lvert \mathbf{c}\cdot(\mathbf{a}\times\mathbf{b})\rvert}{\lvert\mathbf{a}\times\mathbf{b}\rvert}$$ $$-(1)$$

Compute the cross-product:
$$\mathbf{a}\times\mathbf{b}= \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 2&3&4\\ 1&\alpha&1 \end{vmatrix} =(3-4\alpha)\mathbf{i}+2\mathbf{j}+(2\alpha-3)\mathbf{k}.$$

Its squared magnitude is
$$\lvert\mathbf{a}\times\mathbf{b}\rvert^{2}=(3-4\alpha)^{2}+2^{2}+(2\alpha-3)^{2}$$ $$=9-24\alpha+16\alpha^{2}+4+4\alpha^{2}-12\alpha+9$$ $$=20\alpha^{2}-36\alpha+22.$$ $$-(2)$$

Now compute the scalar triple product numerator:
$$\mathbf{AB}\cdot(\mathbf{a}\times\mathbf{b})=(-1,-2,2)\cdot(3-4\alpha,\,2,\,2\alpha-3)$$ $$=(-1)(3-4\alpha)+(-2)(2)+2(2\alpha-3)$$ $$=-3+4\alpha-4+4\alpha-6$$ $$=8\alpha-13.$$ $$-(3)$$

Given shortest distance  $$d=\frac{5}{\sqrt{6}}.$$ Substitute $$-(2)$$ and $$-(3)$$ in $$-(1)$$ and square both sides:

$$\frac{(8\alpha-13)^{2}}{20\alpha^{2}-36\alpha+22}=\frac{25}{6}.$$ Multiplying,

$$6(8\alpha-13)^{2}=25(20\alpha^{2}-36\alpha+22).$$

Expand both sides:
Left: $$6(64\alpha^{2}-208\alpha+169)=384\alpha^{2}-1248\alpha+1014,$$
Right: $$500\alpha^{2}-900\alpha+550.$$

Equate and simplify:
$$384\alpha^{2}-1248\alpha+1014=500\alpha^{2}-900\alpha+550$$
$$0=116\alpha^{2}+348\alpha-464$$
Divide by 4:
$$29\alpha^{2}+87\alpha-116=0.$$ $$-(4)$$

Solve the quadratic $$-(4)$$:
Discriminant $$D=87^{2}-4\cdot29\cdot(-116)=7569+13456=21025=145^{2}.$$
Hence
$$\alpha=\frac{-87\pm145}{2\cdot29}=\frac{-87\pm145}{58}.$$

Therefore
$$\alpha_{1}=1,\qquad \alpha_{2}=-4.$$

The question asks for the sum of all possible values of $$\alpha$$:
$$1+(-4)=-3.$$

Hence the required sum is $$-3$$, corresponding to Option D.