The shortest distance between the lines $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})$ and $\vec{r} = (2\hat{i} + 4\hat{j} + 5\hat{k}) + \mu(3\hat{i} + 4\hat{j} + 5\hat{k})$ is equal to

$\vec{r}_1 = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})$

$\vec{r}_2 = (2\hat{i} + 4\hat{j} + 5\hat{k}) + \mu(3\hat{i} + 4\hat{j} + 5\hat{k})$

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$\vec{a}_1 = (1, 2, 3), \quad \vec{b}_1 = (2, 3, 4)$

$\vec{a}_2 = (2, 4, 5), \quad \vec{b}_2 = (3, 4, 5)$

The shortest distance formula:

$d = \dfrac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}$

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$\vec{a}_2 - \vec{a}_1 = (1, 2, 2)$

$\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix}$

$= (-1)\hat{i} + (2)\hat{j} + (-1)\hat{k}$

$|\vec{b}_1 \times \vec{b}_2| = \sqrt{1 + 4 + 1} = \sqrt{6}$

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$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = (1)(-1) + (2)(2) + (2)(-1)$

$= -1 + 4 - 2 = 1$

$d = \dfrac{|1|}{\sqrt{6}}$

$\boxed{d = \dfrac{1}{\sqrt{6}}}$