Here is a more explicitly computed solution using the complete (not partial) factorization. Using the expansion of Heron's formula and the factorization of $63$, we find that the given Diophantine equation is equivalent to $$(x+y+z)(-x+y+z)(x-y+z)(x+y-z)= 3^2\cdot 7.$$So $x+y+z$ has to be a positive factor of $3^2\cdot 7$. There are $(2+1)(1+1)=6$ such factors, which are $1,3,7,9,21,63$. We will go through these cases backwards. If $x+y+z=63$, then $(-x+y+z,x-y+z,x+y-z)=(1,1,1)$, and if we sum these three values, then we get the contradiction $x+y+z=3$. If $x+y+z=21$, then $(-x+y+z,x-y+z,x+y-z)=(3,1,1)$ or permutations, and if we sum these three values, then we get the contradiction $x+y+z=5$. If $x+y+z=9$, then $(-x+y+z,x-y+z,x+y-z)=(7,1,1)$ or permutations. Summing these two at a time gives the solution $(x,y,z)=\boxed{(1,4,4)}$ and permutations. If $x+y+z=7$, then $(-x+y+z,x-y+z,x+y-z)=(9,1,1)$ or permutations, which are impossible since none of these values should be greater than $x+y+z$. Or $(-x+y+z,x-y+z,x+y-z)=(3,3,1)$ or permutations. Summing these two at a time gives the solution $(x,y,z)=\boxed{(2,2,3)}$ and permutations. If $x+y+z=3$, then $(-x+y+z,x-y+z,x+y-z)=(21,1,1)$ or permutations, or $(-x+y+z,x-y+z,x+y-z)=(1,3,7)$ or permutations. All of these are impossible since none of these values should be greater than $x+y+z$. Finally, $x+y+z=1$ is impossible since that would make it smaller than at least one of the other three factors. Therefore, the solutions are $\boxed{(1,4,4),(4,1,4),(4,4,1),(2,2,3),(2,3,2),(3,2,2)}$.