The shortest distance between the lines $$\vec{r} = \left(\frac{1}{3}\hat{i} + 2\hat{j} + \frac{8}{3}\hat{k}\right) + \lambda(2\hat{i} - 5\hat{j} + 6\hat{k})$$ and $$\vec{r} = \left(-\frac{2}{3}\hat{i} - \frac{1}{3}\hat{k}\right) + \mu(\hat{j} - \hat{k})$$, where $$\lambda, \mu \in \mathbb{R}$$, is :

The two lines are written in vector form as
$$\vec{r} = \vec{a_1} + \lambda \vec{b_1},\qquad \vec{a_1}= \left(\frac13,\,2,\,\frac83\right),\; \vec{b_1}= (2,\,-5,\,6)$$
$$\vec{r} = \vec{a_2} + \mu \vec{b_2},\qquad \vec{a_2}= \left(-\frac23,\,0,\,-\frac13\right),\; \vec{b_2}= (0,\,1,\,-1)$$

The shortest distance $$D$$ between two skew lines
$$\vec{r}=\vec{a_1}+\lambda\vec{b_1}\quad\text{and}\quad \vec{r}=\vec{a_2}+\mu\vec{b_2}$$
is given by
$$D=\frac{\left|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})\right|} {|\vec{b_1}\times\vec{b_2}|}$$

Step 1: Find $$\vec{a_2}-\vec{a_1}$$
$$\vec{a_2}-\vec{a_1}= \left(-\frac23-\frac13,\;0-2,\;-\frac13-\frac83\right)=(-1,\,-2,\,-3)$$

Step 2: Compute $$\vec{b_1}\times\vec{b_2}$$
$$ \vec{b_1}\times\vec{b_2}= \begin{vmatrix} \hat{i}&\hat{j}&\hat{k}\\ 2&-5&6\\ 0&1&-1 \end{vmatrix} =-\hat{i}+2\hat{j}+2\hat{k}=(-1,\,2,\,2) $$

Step 3: Magnitude of the cross product
$$|\vec{b_1}\times\vec{b_2}|=\sqrt{(-1)^2+2^2+2^2}=\sqrt{1+4+4}=3$$

Step 4: Scalar triple product in numerator
$$(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})= (-1,\,-2,\,-3)\cdot(-1,\,2,\,2)= 1-4-6=-9$$
Its absolute value is $$|{-9}|=9$$.

Step 5: Shortest distance
$$D=\frac{9}{3}=3$$

Hence, the shortest distance between the given lines is $$3$$.
Option B which is: $$3$$