Answer: C

Small Hint:

The six fixed numbers sum to $24.8,$ so the mean is an integer only when $x + y + z$ ends in $0.2$

Big Hint:

The range condition gives three cases: $(\min,\max)=(0,7),\ (1,8),$ or $(t,t+7)$ with $0\lt t\lt1$

Solution:

The six fixed numbers sum to $24.8$ and span $[1,7].$ There are three possible arrangements of the overall extremes.

If the extremes are $0$ and $7,$ then $x=0$ and $z\le7.$ The only possible integer means are $3$ and $4,$ requiring $y+z=2.2$ or $11.2.$ The first makes the median $2.2.$ In the second, an integer median forces $y=5,$ hence $z=6.2.$ This gives $(0,5,6.2).$

If the extremes are $1$ and $8,$ then $z=8$ and the integer mean forces $x+y=3.2$ or $12.2.$ The first makes the median $3.2;$ the second has an integer median only for $x=6,\ y=6.2.$ This gives $(6,6.2,8).$

Finally, if both extremes are new, write $x=t,\ z=t+7$ with $0\lt t\lt1.$ The mean must be $4,$ so $y=4.2-2t.$ The median is the fourth number among $1,2.2,3.2,5.2,6.2,7,y;$ it is an integer only when $y=4,$ giving $t=0.1.$ Thus the third triple is $(0.1,4,7.1),$ and there are exactly $3$ in all.

Thus, the correct answer is C.