Equation of parabola in x-y plane 
Given the following equations:
$$x=t^2$$
$$y=t^2$$
$$z=2t+3$$
Show that the graph of the curve $(t^2,t^2,2t+3)$ lives in a plane and that, within that plane, the graph is a parabola.
My attempt, so far:
From the first two equations $x=y$, meaning the curve belongs to the  $x-y=0$ plane.
Plugging the first or second equation into the third one:
$$((z-3)/2)^2=x,\>\>\> ((z-3)/2)^2=y$$
This tells me it is parabola in the zx and zy planes respectively , I guess ??
A matlab plot shows it is , in fact, a parabola in the $x-y=0$  plane. What I am doing wrong and how do I show it?
 A: Rotate the $xyz$-coordinates by 45 degrees around the $z$ axis to the $uvz$-coordinates, which is equivalent to 
$$x=u+v, \>\>\> y=u-v$$
Rewrite the curve in the $uv$-coordinates
$$u+v=t^2, \>\>\>u-v=t^2$$
Then, we have $u=t^2$ and $v=0$. So, the curve lies in the plane $v=0$, with the $uz$ coordinates given by,
$$u = t^2$$
$$z=2t+3$$
Eliminate $t$ to get the standard parabola equation in the plane $v=0$ (or, the plane $x=y$ in the original coordinates)
$$(z-3)^2=4u$$
A: If a curve $u(t)$ lies on a plane, then the plane defined by any three points $A$, $B$, $C$ on the curve must be always the same, i.e. the vector product 
$\vec n=(B-A)\times(C-A)$ must have the same direction for any choice of the points.
In practice we can keep two points fixed and choose, for instance: $A=u(0)=(0,0,3)$, $B=u(1)=(1,1,5)$, $C=u(t)=(t^2,t^2,2t+3)$, to find:
$$
\vec n=(1,1,2)\times(t^2,t^2,2t)=(2t-t^2,2t-t^2,0)=(2t-t^2)(1,1,0).
$$
The direction $\vec n/|\vec n|$ is then ${1\over\sqrt2}(1,1,0)$ and does not depend on $t$: hence the curve lies on a plane whose normal unit vector is ${1\over\sqrt2}(1,1,0)$.
