The shortest distance between the lines $\vec{r} = \hat{i} + \hat{j} + \lambda(2\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = 2\hat{i} + \hat{j} - \hat{k} + \mu(4\hat{i} - 2\hat{j} + 2\hat{k})$ is

Line 1: $\vec{r} = \hat{i} + \hat{j} + \lambda(2\hat{i} - \hat{j} + \hat{k})$

Fixed point: $\vec{a_1} = \hat{i} + \hat{j} + 0\hat{k}$ → $(1, 1, 0)$

Direction vector: $\vec{b_1} = 2\hat{i} - \hat{j} + \hat{k}$ → $(2, -1, 1)$

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

Fixed point: $\vec{a_2} = 2\hat{i} + \hat{j} - \hat{k}$ → $(2, 1, -1)$

Direction vector: $\vec{b_2} = 4\hat{i} - 2\hat{j} + 2\hat{k}$ → $(4, -2, 2)$

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Notice that $\vec{b_2} = 2\vec{b_1}$:

$\vec{b_2} = (4, -2, 2) = 2(2, -1, 1) = 2\vec{b_1}$

The lines are parallel to each other.

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For parallel lines, the shortest distance is:

$d = \frac{|\vec{b_1} \times (\vec{a_2} - \vec{a_1})|}{|\vec{b_1}|}$

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Calculate $\vec{a_2} - \vec{a_1}$:

$\vec{a_2} - \vec{a_1} = (2, 1, -1) - (1, 1, 0)$

$\vec{a_2} - \vec{a_1} = (1, 0, -1)$

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Calculate the cross product $\vec{b_1} \times (\vec{a_2} - \vec{a_1})$:

$\vec{b_1} \times (\vec{a_2} - \vec{a_1}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 0 & -1 \end{vmatrix}$

$\hat{i}$ component: $(-1)(-1) - (1)(0) = 1$

$\hat{j}$ component: $-[(2)(-1) - (1)(1)] = -[-3] = 3$

$\hat{k}$ component: $(2)(0) - (-1)(1) = 1$

Therefore: $\vec{b_1} \times (\vec{a_2} - \vec{a_1}) = \hat{i} + 3\hat{j} + \hat{k}$ → $(1, 3, 1)$

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Calculate the magnitude of the cross product:

$|\vec{b_1} \times (\vec{a_2} - \vec{a_1})| = \sqrt{1^2 + 3^2 + 1^2}$

$|\vec{b_1} \times (\vec{a_2} - \vec{a_1})| = \sqrt{11}$

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Calculate the magnitude of $\vec{b_1}$:

$|\vec{b_1}| = \sqrt{2^2 + (-1)^2 + 1^2}$

$|\vec{b_1}| = \sqrt{6}$

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Calculate the shortest distance:

$d = \frac{\sqrt{11}}{\sqrt{6}}$

$d = \frac{\sqrt{11}}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}$

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

Therefore, the shortest distance between the lines is $\frac{\sqrt{66}}{6}$.