What will be printed?

Main.java

import java.nio.file.Path;
import java.nio.file.Paths;

/**
 * Solution: https://ideone.com/qcHw1f
 */
class Main {

  public static void main(String[] args) {
    Path path = Paths.get("/");
    foo(path);
  }

  private static void foo(Path... path) {
    System.out.println("Vararg");
  }

  private static void foo(Iterable<Path> path) {
    System.out.println("Iterable");
  }

}

It will print "Iterable" because Path implements Iterable on path

It will print "Iterable", but the reason is not because Path implements Iterable. See what happens with:

foo(Arrays. <Path>asList(path));
foo(new Path[] {path});

or, if we change the signature of foo(Path... path) to foo(Path path).

So, must have something to do with the varargs, but I don't know why...

Well, Java probably first tries to find a method suitable for the args, and finds foo(Iterable path), because Path implements Iterable.
The lookup for the method with the varargs is just left as a second option, if no method had been found in the first lookup.