Answer: A

Solution:

Because $\text{lcm}(x,y)=2^3\cdot3^2$ and $\text{lcm}(x,z)=2^3\cdot3\cdot5^2,$ the factor $5^2$ divides $z$ while neither $x$ nor $y$ is divisible by $5.$ Also $3^2$ divides $y,$ while neither $x$ nor $z$ is divisible by $3^2,$ and $x$ must have the factor $2^3.$

Writing $x=2^3\cdot3^{\,j},$ $y=2^{\,k}\cdot3^2,$ and $z=2^{\,m}\cdot3^{\,n}\cdot5^2,$ the lcm conditions require $\max(j,n)=1$ and $\max(k,m)=2.$ The first pair can be $(1,0),(0,1),(1,1),$ and the second can be $(2,0),(2,1),(2,2),(0,2),(1,2).$ Thus there are $3\cdot5=15$ ordered triples.

Thus, the correct answer is A.