Answer: C

Solution:

The six fixed numbers total $24.8.$ If the mean is the integer $k,$ then $x+y+z=9k-24.8.$ Because the fixed entries already run from $1$ to $7,$ the range condition gives three cases.

If $z\le7,$ then $x=0.$ The bounds on $y+z$ force $k=3$ or $4.$ For $k=3,$ the median is $2.2;$ for $k=4,$ we have $y+z=11.2,$ so $y\ge4.2$ and the median is integral only when $y=5,$ giving $(x,y,z)=(0,5,6.2).$

If $x\ge1,$ then $z=8.$ Here $k=4$ gives median $3.2.$ For $k=5,$ we have $x+y=12.2;$ the median is integral only when $x=6,$ giving $(6,6.2,8).$

The remaining case has $0<x<1$ and $z=x+7.$ The total lies between $31.8$ and $41.8,$ so $k=4$ and $y=4.2-2x.$ The median can be an integer only when $y=4,$ giving $(0.1,4,7.1).$ Hence exactly $3$ ordered triples work. Thus, C is the correct answer.