Solution: There are three different routes from A to C 1, that is, three different expansion modes.

First, the surface BCC 1B 1 is expanded along BB 1 to be in the same plane as the surface ABA 1B 1.

∴AC 1=√(3？ +( 1+2)? )=3√2

Second, along A 1B 1, expand the plane A 1b 1d 1 into a plane parallel to the plane ABA1b1* *+0 * *.

∴AC 1=√( 1+(2+3)？ )=√26

The third type: along a1d1d1d1,expand the plane A 1A 1 * * into a parallel plane.

∴AC 1=√(2？ +( 1+3)? )=2√5

To sum up, the shortest distance from AC 1 to the surface of a cuboid is 3√2.