$100x+10y+z=x+y+z+xy+yz+zx+xyz$ $9(x+y)+90x=z(x+y)+xy(1+z)$ $(9-z)(x+y)=x[y(z+1)-90]$ $9 \ge y,z $ $\implies$ $LHS \ge 0$ and $0 \ge RHS$ The only solution is $LHS=RHS=0$ $\implies$ $y=z=9.$ $x$ can take any value from $1$ to $9$ The solutions to the equation are $199, 299, 399 ... 999.$