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 $\vec{r} = (2\hat{i} + 4\hat{j} + 5\hat{k}) + \mu(4\hat{i} + 6\hat{j} + 8\hat{k})$ is equal to

Line 1: $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})$

Point on line: $\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}$

Direction vector: $\vec{b_1} = 2\hat{i} + 3\hat{j} + 4\hat{k}$

Line 2: $\vec{r} = (2\hat{i} + 4\hat{j} + 5\hat{k}) + \mu(4\hat{i} + 6\hat{j} + 8\hat{k})$

Point on line: $\vec{a_2} = 2\hat{i} + 4\hat{j} + 5\hat{k}$

Direction vector: $\vec{b_2} = 4\hat{i} + 6\hat{j} + 8\hat{k}$

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Notice: $\vec{b_2} = 4\hat{i} + 6\hat{j} + 8\hat{k}$

$\vec{b_2} = 2(2\hat{i} + 3\hat{j} + 4\hat{k})$

$\vec{b_2} = 2\vec{b_1}$

Since $\vec{b_2} = 2\vec{b_1}$, the lines are parallel.

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For parallel lines, the shortest distance is:

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

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$\vec{a_2} - \vec{a_1} = (2-1)\hat{i} + (4-2)\hat{j} + (5-3)\hat{k}$

$\vec{a_2} - \vec{a_1} = \hat{i} + 2\hat{j} + 2\hat{k}$

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$(\vec{a_2} - \vec{a_1}) \times \vec{b_1} = (\hat{i} + 2\hat{j} + 2\hat{k}) \times (2\hat{i} + 3\hat{j} + 4\hat{k})$

$= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 3 & 4 \end{vmatrix}$

$\hat{i}$ component: $(2)(4) - (2)(3) = 8 - 6 = 2$

$\hat{j}$ component: $-[(1)(4) - (2)(2)] = -(4 - 4) = 0$

$\hat{k}$ component: $(1)(3) - (2)(2) = 3 - 4 = -1$

$(\vec{a_2} - \vec{a_1}) \times \vec{b_1} = 2\hat{i} + 0\hat{j} - \hat{k}$

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$|(\vec{a_2} - \vec{a_1}) \times \vec{b_1}| = \sqrt{(2)^2 + (0)^2 + (-1)^2}$

$= \sqrt{4 + 0 + 1}$

$= \sqrt{5}$

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$|\vec{b_1}| = \sqrt{(2)^2 + (3)^2 + (4)^2}$

$= \sqrt{4 + 9 + 16}$

$= \sqrt{29}$

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

$d = \frac{\sqrt{5}}{\sqrt{29}}$

$d = \sqrt{\frac{5}{29}}$

Therefore, the shortest distance between the lines is $\sqrt{\frac{5}{29}}$.