Let the values of $$\lambda$$ for which the shortest distance between the lines $$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$$ and $$\frac{x-\lambda}{3} = \frac{y-4}{4} = \frac{z-5}{5}$$ is $$\frac{1}{\sqrt{6}}$$ be $$\lambda_1$$ and $$\lambda_2$$. Then the radius of the circle passing through the points $$(0, 0)$$, $$(\lambda_1, \lambda_2)$$ and $$(\lambda_2, \lambda_1)$$ is :

The given lines can be rewritten in the symmetric-parametric form:
$$L_1 : \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=t \;\;\Longrightarrow\;\; \mathbf{r}_1=(1,2,3)+t\,(2,3,4)$$ $$L_2 : \frac{x-\lambda}{3}=\frac{y-4}{4}=\frac{z-5}{5}=s \;\;\Longrightarrow\;\; \mathbf{r}_2=(\lambda,4,5)+s\,(3,4,5)$$

For two skew lines $$\mathbf{r}=\mathbf{a}_1+t\,\mathbf{b}_1$$ and $$\mathbf{r}=\mathbf{a}_2+s\,\mathbf{b}_2$$, the shortest distance $$D$$ is given by
$$D=\frac{\left|(\mathbf{a}_2-\mathbf{a}_1)\,\cdot\,(\mathbf{b}_1\times\mathbf{b}_2)\right|}{\left|\mathbf{b}_1\times\mathbf{b}_2\right|}$$

Here
$$\mathbf{a}_1=(1,\,2,\,3),\quad \mathbf{b}_1=(2,\,3,\,4)$$ $$\mathbf{a}_2=(\lambda,\,4,\,5),\quad \mathbf{b}_2=(3,\,4,\,5)$$

First find $$\mathbf{b}_1\times\mathbf{b}_2$$:
$$\mathbf{b}_1\times\mathbf{b}_2= \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 2&3&4\\ 3&4&5 \end{vmatrix} =\bigl(3\cdot5-4\cdot4,\;4\cdot3-2\cdot5,\;2\cdot4-3\cdot3\bigr) =(-1,\,2,\,-1)$$ $$\left|\mathbf{b}_1\times\mathbf{b}_2\right|=\sqrt{(-1)^2+2^2+(-1)^2}=\sqrt6$$

Next evaluate $$(\mathbf{a}_2-\mathbf{a}_1)\cdot(\mathbf{b}_1\times\mathbf{b}_2)$$:
$$\mathbf{a}_2-\mathbf{a}_1=(\lambda-1,\,4-2,\,5-3)=(\lambda-1,\,2,\,2)$$ $$\bigl(\lambda-1,\,2,\,2\bigr)\cdot(-1,\,2,\,-1)=-(\lambda-1)+4-2=-\lambda+3$$ Hence $$\left|(\mathbf{a}_2-\mathbf{a}_1)\cdot(\mathbf{b}_1\times\mathbf{b}_2)\right|=\left|\,\lambda-3\,\right|$$

The shortest distance is given to be $$\dfrac1{\sqrt6}$$, therefore
$$\frac{\left|\lambda-3\right|}{\sqrt6}=\frac1{\sqrt6}\;\;\Longrightarrow\;\;\left|\lambda-3\right|=1$$ Case 1: $$\lambda-3=1\;\Longrightarrow\;\lambda_1=4$$
Case 2: $$\lambda-3=-1\;\Longrightarrow\;\lambda_2=2$$

Thus $$\lambda_1=2,\;\lambda_2=4$$ (order is immaterial).

The circle must pass through the three points
$$A\,(0,0),\quad B\,(\lambda_1,\lambda_2)=(2,4),\quad C\,(\lambda_2,\lambda_1)=(4,2)$$

Compute the side lengths of $$\triangle ABC$$:
$$AB=\sqrt{(2-0)^2+(4-0)^2}=2\sqrt5$$ $$AC=\sqrt{(4-0)^2+(2-0)^2}=2\sqrt5$$ $$BC=\sqrt{(4-2)^2+(2-4)^2}=2\sqrt2$$

Since $$AB=AC$$, the triangle is isosceles. The area $$\Delta$$ can be obtained using the determinant method in 2-D:
$$\Delta=\frac12\left|\,\begin{vmatrix}2&4\\4&2\end{vmatrix}\right| =\frac12\left|\,2\cdot2-4\cdot4\,\right| =\frac12\left|\,4-16\,\right|=6$$

For a triangle with sides $$a,b,c$$ and area $$\Delta$$, the circum-radius $$R$$ is
$$R=\frac{abc}{4\Delta}$$ Taking $$a=BC=2\sqrt2,\; b=AC=2\sqrt5,\; c=AB=2\sqrt5$$:
$$\begin{aligned} R&=\frac{(2\sqrt2)(2\sqrt5)(2\sqrt5)}{4\times6}\\ &=\frac{8\sqrt2\,(5)}{24}\\ &=\frac{40\sqrt2}{24}=\frac{5\sqrt2}{3} \end{aligned}$$

Hence the radius of the required circle is $$\dfrac{5\sqrt2}{3}$$, which corresponds to Option A.