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

The given lines are

$$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda}$$

and

$$\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}$$

A point on the first line is

$$A_1=(1,2,3)$$

and its direction vector is

$$\vec b_1=2\hat i+3\hat j+\lambda\hat k$$

A point on the second line is

$$A_2=(2,4,5)$$

and its direction vector is

$$\vec b_2=\hat i+4\hat j+5\hat k$$

The shortest distance between two skew lines is

$$\text{S.D.}=\frac{|(\vec{A_2A_1})\cdot(\vec b_1\times\vec b_2)|}{|\vec b_1\times\vec b_2|}$$

Given,

$$\text{S.D.}=\frac1{\sqrt3}$$

Now,

$$\vec{A_2A_1}=(2-1)\hat i+(4-2)\hat j+(5-3)\hat k$$

$$=\hat i+2\hat j+2\hat k$$

Next,

$$\vec b_1\times\vec b_2= \begin{vmatrix} \hat i&\hat j&\hat k\\ 2&3&\lambda\\ 1&4&5 \end{vmatrix}$$

$$=\hat i(15-4\lambda)-\hat j(10-\lambda)+\hat k(8-3)$$

$$=(15-4\lambda)\hat i+(\lambda-10)\hat j+5\hat k$$

Therefore,

$$|\vec b_1\times\vec b_2| =\sqrt{(15-4\lambda)^2+(\lambda-10)^2+25}$$

Also,

$$(\vec{A_2A_1})\cdot(\vec b_1\times\vec b_2)$$

$$=(1,2,2)\cdot(15-4\lambda,\lambda-10,5)$$

$$=15-4\lambda+2\lambda-20+10$$

$$=5-2\lambda$$

Hence,

$$\frac{|5-2\lambda|}{\sqrt{(15-4\lambda)^2+(\lambda-10)^2+25}}=\frac1{\sqrt3}$$

Squaring both sides,

$$3(5-2\lambda)^2=(15-4\lambda)^2+(\lambda-10)^2+25$$

Expanding,

$$3(25-20\lambda+4\lambda^2)=225-120\lambda+16\lambda^2+\lambda^2-20\lambda+100+25$$

$$75-60\lambda+12\lambda^2=17\lambda^2-140\lambda+350$$

$$5\lambda^2-80\lambda+275=0$$

$$\lambda^2-16\lambda+55=0$$

$$(\lambda-5)(\lambda-11)=0$$

Thus,

$$\lambda=5,\ 11$$

Therefore, the sum of all possible values of $$\lambda$$ is

$$5+11=16$$

Hence, $$\boxed{16}$$.