Three Dice minimum value

A lower bound is

$73$: when getting the sum to be $216$, the average value must be $72$. However, this would require three of the numbers to be the same. Hence at least one number must be $73$ or bigger.

An upper bound is

$77$: achieved by the dice $\{ 72, 73, 74, 75, 76, 77 \}, \{ 40, 46, 52, 58, 64, 70 \}$ and $\{ -111, -75, -39, -3, 33, 69 \}$. These dice come from taking the units, sixes and thirty-sixes in the base-six representation, which represent the integers between $0$ and $215$ uniquely. We then shift the units by $72$, the sixes by $40$ and the thirty-sixes by $-111$ to get the numbers from $1$ to $216$ without repeating any faces of the dice.

The smallest number I could find was 75. I think the way to prove it is to prove that the only way to get 216 distinct consecutive numbers by summing 3 three dice is for each die to be of the form, $a + n*6^b$

Here is what I did:

I started with these 3 sequences which are able to reproduce the number 1-216. Basically this is the first three digits of a base 6 representation. Each row represents one die.

1, 2, 3, 4, 5, 6
0, 6, 12, 18, 24, 30
0, 36, 72, 108, 144, 180

Then I added or subtracted from each row (die). The numbers I used were +68 +37 -105. Since the sum of these 3 numbers is 0, the new set is still able to produce the numbers 1-216:

Here are the resulting 3 dice:

69, 70, 71, 72, 73, 74
37, 43, 49, 55, 61, 67
-105, -69, -33, 3, 39, 75

I was able to find an option where the largest number was 73, but it required a repeat. The only way I could find where there were no repeats was to have every number in the first row (die) greater than every number in the second row (die)

Edit: This is wrong, I missed the part about the numbers needing to be distinct.

The largest die must be

72

or more. Otherwise, the largest number you could roll would be

71 + 71 + 71 = 213 < 216.

Here's a labeling which achieves that minimum:

67, 68, 69, 70, 71, 72
42, 48, 54, 60, 66, 72
-108, -72, -36, 0, 36, 72