Here is a partial solution.
It is easy to reduce the equation to xy+xz+yz=a.
It is also easy to see that 2 cannot divide a.
Let x=(4k^2+2k+a), y=8k^2+2k+2a, z=8k^2+6k+2a+1.
We compute that xy+xz+yz=a for any k.
On the other hand, taking k==1 mod a, we see that x==2 mod a, y==6 mod a, and z==3 mod a.
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Note that as k increases (for large enough k), x,y,z also increase in magnitude.

Thus, this solution works for any a not divisible by 2 or 3.