The shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-2}{4} = \frac{y-4}{6} = \frac{z-5}{8}$ is equal to

Line 1: $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$

This line passes through point $A_1 = (1, 2, 3)$ and has direction vector $\vec{b_1} = (2, 3, 4)$

Line 2: $\frac{x-2}{4} = \frac{y-4}{6} = \frac{z-5}{8}$

This line passes through point $A_2 = (2, 4, 5)$ and has direction vector $\vec{b_2} = (4, 6, 8)$

---

Notice that $\vec{b_2} = (4, 6, 8) = 2 \times (2, 3, 4) = 2\vec{b_1}$

The lines are parallel since their direction vectors are proportional.

---

For parallel lines, the shortest distance is:

$d = \frac{|\vec{A_1A_2} \times \vec{b_1}|}{|\vec{b_1}|}$

The vector $\vec{A_1A_2}$ is:

$\vec{A_1A_2} = A_2 - A_1$

$\vec{A_1A_2} = (2-1, 4-2, 5-3)$

$\vec{A_1A_2} = (1, 2, 2)$

---

The cross product $(1, 2, 2) \times (2, 3, 4)$ is:

$\vec{A_1A_2} \times \vec{b_1} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 2 & 2 \\ 2 & 3 & 4 \end{vmatrix}$

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

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

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

Therefore $\vec{A_1A_2} \times \vec{b_1} = (2, 0, -1)$

---

The magnitude of the cross product is:

$|\vec{A_1A_2} \times \vec{b_1}| = \sqrt{2^2 + 0^2 + (-1)^2}$

$|\vec{A_1A_2} \times \vec{b_1}| = \sqrt{4 + 0 + 1}$

$|\vec{A_1A_2} \times \vec{b_1}| = \sqrt{5}$

---

The magnitude of $\vec{b_1}$ is:

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

$|\vec{b_1}| = \sqrt{4 + 9 + 16}$

$|\vec{b_1}| = \sqrt{29}$

---

The shortest distance is:

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

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