Here is the problem bothering me.
Let the sphere $S :x^2+(y-2)^2+(z-3)^2=1$
Put the sphere's tangent line $l s.t.$ passing through a point $C(0,0,c)$ on $z$-axis.
Say the intersection point $p$ between $l $and $xy$-plain.
Find the points $C(0,0,c)$ whose locus of the $p$ is a parabola on $xy$-plain.
The set of lines $l$ tangent to a sphere, passing through a fixed point $C$, is nothing but a cone. It s well known that the intersection of a cone and a plane is a conic, in particular it may be an ellipse, a parabola, or an hyperbola.
Now, the resulting intersection is a parabola if and only if exactly one of the lines $l$ is parallel to the plane. In the case of the $xy$-plane, it means that the line $l$ is contined in some plane of the form $z=c$, where $c$ is a constant.
Such a plane is tangent to the sphere if and only if $c=3 \pm 1$, so these are the two values of $c$ which solve your problem. A picture will help to fill the details. Here the $x$ axis represents the $xy$-plane, the circle represents the sphere, and the two lines represent the two tangent planes.