The shortest distance between the lines $\vec r= \hat i+\hat j+λ(2\hat i-\hat j+\hat k)$ and $\vec r = 2\hat i+\hat j-\hat k+μ(4\hat i-2\hat j + 2\hat k)$ is

$\frac{\sqrt{66}}{6}$

The correct answer is Option (1) → $\frac{\sqrt{66}}{6}$

Given lines:

$\vec{r}_1 = \hat{i} + \hat{j} + \lambda(2\hat{i} - \hat{j} + \hat{k})$

$\vec{r}_2 = 2\hat{i} + \hat{j} - \hat{k} + \mu(4\hat{i} - 2\hat{j} + 2\hat{k})$

Direction ratios:

Line 1 → $(2, -1, 1)$

Line 2 → $(4, -2, 2)$ = $2(2, -1, 1)$ → lines are parallel.

Position vectors of given points:

$\vec{a}_1 = (1, 1, 0)$

$\vec{a}_2 = (2, 1, -1)$

Vector joining the points:

$\vec{a}_2 - \vec{a}_1 = (1, 0, -1)$

For parallel lines, shortest distance:

$D = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}_1|}{|\vec{b}_1|}$

$(\vec{a}_2 - \vec{a}_1) \times \vec{b}_1 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & -1 \\ 2 & -1 & 1 \end{vmatrix} = (-1)\hat{i} - 3\hat{j} - 1\hat{k}$

$|(\vec{a}_2 - \vec{a}_1) \times \vec{b}_1| = \sqrt{1 + 9 + 1} = \sqrt{11}$

$|\vec{b}_1| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{6}$

Therefore, $D = \frac{\sqrt{11}}{\sqrt{6}} = \sqrt{\frac{11}{6}}$

Final Answer: $\sqrt{\frac{11}{6}}$