119. What is the shortest distance on the surface of the given right parallelepiped (edges: $a>b>c$) between one of the vertices and the opposite vertex (the opposite vertex to $M$ is considered to be the vertex $N$, where the three faces parallel to the faces meeting at vertex $M$ meet)?
119. Unfold the surface of the parallelepiped onto a plane by cutting along some edges; this can be done in three different ways; by then connecting the new positions of point $\boldsymbol{M}$ with $N$, we obtain three straight segments, the shortest of which will be the required shortest distance; the lengths of these segments are:

$$
\sqrt{a^{2}+(b+c)^{2}}, \quad \sqrt{b^{c}+(c+a)^{2}}, \sqrt{c^{2}+(a+b)^{2}}
$$

since $a>b>c$, the shortest segment will be

$$
\sqrt{a^{2}+(b+c)^{2}}
$$