7. Real numbers $x, y, z$ satisfy
$$
x^{2}+2 y=7, y^{2}+4 z=-7, z^{2}+6 x=-14 \text {. }
$$

Then $x^{2}+y^{2}+z^{2}$ equals $\qquad$
$$
\begin{array}{l}
\left(x^{2}+6 x\right)+\left(y^{2}+2 y\right)+\left(z^{2}+4 z\right)=-14 . \\
\text { Therefore, }(x+3)^{2}+(y+1)^{2}+(z+2)^{2}=0 .
\end{array}
$$

So, $x=-3, y=-1, z=-2$.