$x$ and $y$ and $z$

With $y = 4 - x,$ the second equation becomes

$x(4 - x) + 6z = z^{2} + 13 \Longrightarrow z^{2} - 6z + (13 + x^{2} - 4x) = 0.$

Using the quadratic equation, we see that the possible values for $z$ are

$z = \dfrac{6 \pm \sqrt{36 - 4(13 + x^{2} - 4x)}}{2} = \dfrac{6 \pm \sqrt{-16 - 4x^{2} + 16x}}{2} =$

$3 \pm \sqrt{-(x^{2} - 4x + 4)} = 3 \pm \sqrt{-(x - 2)^{2}}.$

Since $-(x - 2)^{2} \le 0$ for real $x,$ for $z$ to also be real we will require that $x = 2,$ in which case $z = 3$ and $y = 4 - x = 2.$

Thus $x + y + z = 2 + 2 + 3 = \boxed{7}$ is the unique solution over $\mathbb{R}.$

x+y=4 Keeping only integer solutions in mind we get five pairs of possible solutions of this equation i.e. (0,4),(1,3),(2,2),(3,1),(4,0).

Now from the second equation we get

   xy+6z          =   z^2+13

or, z^2-6z+9 = xy-4 or, (z-3)^2 = (xy-4)

Now to get a reliable solution from the given equation, we put x=2 and y=2 as the most appropiate ones through trial and error method.

Hence we get,

  (z-3)^2 = 0

or, z=3

Hence -> (x+y+z)=(4+3)= 7 (Answer)