For two skew lines, the shortest distance formula is:
$$d = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|}$$
From the given lines, identify:
- Point on Line 1: $$\vec{a_1} = \langle 1, 2, 3 \rangle$$, Direction: $$\vec{b_1} = \langle 2, 3, 4 \rangle$$
- Point on Line 2: $$\vec{a_2} = \langle 4, 5, 6 \rangle$$, Direction: $$\vec{b_2} = \langle 1, 2, 2 \rangle$$
Calculate the difference vector:
$$\vec{a_2} - \vec{a_1} = \langle 3, 3, 3 \rangle$$
Calculate the cross product $$\vec{b_1} \times \vec{b_2}$$:
$$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 3 & 4 \\ 1 & 2 & 2 \end{vmatrix}$$
$$= \vec{i}(3 \cdot 2 - 4 \cdot 2) - \vec{j}(2 \cdot 2 - 4 \cdot 1) + \vec{k}(2 \cdot 2 - 3 \cdot 1)$$
$$= \vec{i}(-2) - \vec{j}(0) + \vec{k}(1) = \langle -2, 0, 1 \rangle$$
Calculate the dot product:
$$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = \langle 3, 3, 3 \rangle \cdot \langle -2, 0, 1 \rangle = -6 + 0 + 3 = -3$$
Calculate the magnitude of the cross product:
$$|\vec{b_1} \times \vec{b_2}| = \sqrt{4 + 0 + 1} = \sqrt{5}$$
Therefore, the shortest distance is:
$$d = \frac{|-3|}{\sqrt{5}} = \frac{3}{\sqrt{5}} = \frac{3\sqrt{5}}{5}$$
Express in the form $$\frac{\sqrt{n}}{m}$$:
$$\frac{3\sqrt{5}}{5} = \frac{\sqrt{9 \cdot 5}}{5} = \frac{\sqrt{45}}{5}$$
So $$n = 45$$ and $$m = 5$$
Therefore: $$n + m = 45 + 5 = 50$$
$$\boxed{50}$$
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