Minimize the longest King chain on a 6x6 ternary grid

Question

This puzzle is an extension of this one: Minimize the longest King chain on a 5x5 binary board

Given a grid filled with numbers, we define a King chain to be a path on the grid such that:

1. The path can be traversed with chess King's moves (moving to one of 8 adjacent cells at a time),
2. The numbers on the path are all equal, and
3. The path does not visit any cell more than once.

The length of a King chain is the number of cells visited by the King.

Can you find a 6x6 grid filled with zeros, ones and twos, such that every pair of cells with the same number are connected via some King chain, and the longest King chain has length 7?

Answer 1

EDIT: All solutions up to mirror symmetry, rotation, colour permutation found by brute force:

End of EDIT.

One possibility (I write X,O,+ for 0,1,2):

    + X + O X X
    O + X O X O
    + O X X O O
    + X O O X X
    + O + X O X
    O + O X O O
 

A bit on how this was found:

To cover as much as possible with one path as short as possible we want it to branch:

Here is one promising pattern (left core pattern, right with example set of subbranches)

    . . . . . .     . . . O . O
    O . . . . O     O . . O . O
    . O . . O .     . O . . O .
    . . O O . .     . . O O . .
    . O . . O .     . O . . O O
    O . . . . O     O . O . . O
 

We can neatly fit two of those:

    X . . . . X
    O X . . X O
    . O X X O .
    . X O O X .
    X O . . O X
    O . . . . O
 

Now the centre is blocked we can reasonably expect the third colour to cover maybe half the grid but not all. The other half we need to cover using only X and O. This can be done symmetrically:

    X . X O X X
    O X . O X O
    . O X X O O
    . X O O X X
    X O . X O X
    O . O X O O
 

It remains to connect the leftover area. This can be done by small adjustments that break symmetry

    . X . O X X
    O . X O X O
    . O X X O O
    . X O O X X
    . O . X O X
    O . O X O O
 

Answer 2

It feels like this can be improved but the best I've been able to do so far is

8

As follows

2 2 2 2 2 2
2 0 1 0 1 2
1 0 1 0 1 0
1 0 1 0 1 0
0 1 0 1 0 1
0 1 0 1 0 1