We have three cases:
**Case I:** $x,y,z\
e0$
There are ${15-1\choose 3-1}$ ways to distribute the $15$. But since $x,y,z$ can be negative, we have $2^3$ ways to distribute the negative sign. So this gives us a total of $8\cdot{15-1\choose 3-1}$ ways.
**Case II:** Only one out of $x,y,z$ is $0$.
There are three ways to make any one of $x,y,z$ equal to $ 0$. For example, let $x=0$. Now we need to distribute the 15 between $y,z$. This can be done in ${15-1\choose2-1}$. Now we can also make any of $y,z$negative. This can be done in $2^2$ ways. So we get a total of $2^2\cdot3\cdot{15-1\choose2-1}$ ways.
**Case III:** Two out of $x,y,z $ are $ 0$.
The zeroes can be distributed in $3$ ways. Let us say $x,y=0$. Now $z$ can be $15,-15$. So we have 6 ways to do this.
Adding 'em up, we get a total of $728+168+6=\boxed{902}$ ways.