2. Adding the first equation to the other two, we get:

$$
x^{2}+2 x+y^{2}+4 y+z^{2}+6 z=-14
$$

Next, we complete the square for each variable. As a result, we obtain:

$$
(x+1)^{2}+(y+2)^{2}+(z+3)^{2}=0
$$

Therefore, $\mathrm{x}=-1, y=-2, z=-3$ is the only possible solution. It remains to verify this by substituting it into the second and third equations of the system.
Answer: $x=-1, y=-2, z=-3$.