Answer
Inequalities represent the set of all points in a 3 -space that are out of the cylinder $16=(-2+z)^{2}+(3+y)^{2}$
Work Step by Step
We complete squares in the given inequality
\[
\begin{aligned}
6 y-4 z+z^{2}+y^{2}>3 & \Rightarrow\left(-4 z+z^{2}+4\right)+\left(6 y+y^{2}+9\right)>3+9+4 \\
& \Rightarrow(-2+z)^{2}+(3+y)^{2}>16
\end{aligned}
\]
Notice that
\[
16=(-2+z)^{2}+(3+y)^{2}
\]
It's a cylinder whose projection is at the level $yz$ in which the circle
\[
16=(-2+z)^{2}+(3+y)^{2}
\]
is centered at (0,-3,2) of radius $4=r $. And therefore the inequality represents the set of all points in 3 -space that are out of the cylinder $16=(-2+z)^{2}+(3+y)^{2}$
