Find where space curves intersect

r1[t_]={t,1-t,3+(t^2)};
r2[s_]={3-s,s-2,(s^2)};

Manipulate[Show[
{ParametricPlot3D[r1[s],{s,-5,sMax}],
ParametricPlot3D[r2[t],{t,-5,tMax}]},
PlotRange->{{-6,8},{-6,8},{0,25}},AxesLabel->{x,y,z},Ticks->False],

(*Manipulate Controllers*)
 Style["Intersection of parametric curves",Bold,12],(*Title*)
{{sMax,0,"s"},-4.9,5,0.1},
{{tMax,0,"t"},-4.9,5,0.1},

(*Notes*)
Delimiter,
Style["",Plain,8],
Style["An intersection takes place when",Plain,8],
Style["the curves share an (x,y,z) point.",Plain,8],
Style["Note: This does not necessarily take",Plain,8],
Style["place when t=s",Plain,8],
ControlPlacement->Left
]

Notice that the curves live in the same plane.  This is why setting the x or y coordinates equal gives an infinite number of solutions.  Setting the z coordinates equal gives 3+(t^2)=(s^2).  Now, from the x coordinates, we have that t=3-s.  Squaring both sides gives, (t^2)=(((3-s))^2)=(s^2)-3 (from the equivalence of the z coordinates).  So, (s^2)-3=(s^2)-6s+9->6s=12->s=2.  If s=2, then t=1.  Now we can find the point of intersection (1,0,4) by using the known values for t or s.

To find the angle at the point of intersection, find the tangent vector of each position vector at the appropriate t and s and then use the dot product to find the angle.