Answer
$(0,0,0)$ and $(1,0,1)$
Work Step by Step
The parametric equation from $r(t)= ti +(2t-t^2) k$ is given as:
$x=t; y=0, z=2t-t^2$
We need to plug these parametric equations in to the given paraboloid equation as: $z=x^2+y^2$
$2t-t^2=t^2+0^2$
or, $2t-2t^2=0$
or, $2t(1-t) =0 \implies t=0,1$
Now, the values of $t$ for the point of intersection are:
$x=0; y=0$ and $z=2(0)-(0)^2=0$
and
$x=1; y=0$ and $z=2(1)-(1)^2=1$
Thus, the the points of intersection are: $(0,0,0)$ and $(1,0,1)$