First we find out how many groups of \(3\) players out of \(10\) can be formed and then decide how many ways we can give each of these groups of three players, Gold, Silver and Bronze prizes.

Now, since John always gets a prize, he must always be in the group of \(3\) players.

So we need to find out the number of ways we can select \(2\) players out of \(9\) players

Number of groups \(= \frac{9!}{2! \times 7!} = \frac{9 \times 8 \times 7!}{2 \times 7!} = \frac{9 \times 8}{2} = 36\)

There are \(36\) possible groups of \(3\) players each.

Within each group, we can distribute the Gold, Silver and Bronze prizes to the three players in \(3!\) or \(6\) different ways.

For \(36\) groups, we can distribute the prizes in \(36 \times 6 = 216\) ways.