How do I find a tangent plane without a specified point? I was having a problem finding the points on $z=3x^2 - 4y^2$ where vector $n=<3,2,2>$ is normal to the tangent plane. 
How do we calculate the tangent plane equation without a specific point to calculate it at? 
I also had an idea to take the cross product of $2$ vectors in the plane and somehow compare it to the $n$ vector but I don't know exactly how to do this. Thank you for any help!
 A: Let $f(x,y,z)=3x^2-4y^2-z$. Then your surface is $\bigl\{(x,y,z)\in\mathbb{R}^3\,|\,f(x,y,z)=0\bigr\}$. You are after the points $(x,y,z)$ in that surface such that $\nabla f(x,y,z)$ is a multiple of $(3,2,2)$. So, solve the system$$\left\{\begin{array}{l}6x=3\lambda\\-8y=2\lambda\\-1=2\lambda\\3x^2-4y^2-z=0.\end{array}\right.$$
A: Let $ f(x,y,z)=3x^2 - 4y^2 - z$ then the normal vector at $p_0(x_0,y_0,z_0)$ is
$$\nabla f(p)=(f_x,f_y,f_z)_p$$
or
$$\nabla f(p)=(6x_0,-8y_0,-1)$$
then
$$\dfrac{\nabla f(p)}{|\nabla f(p)|}=\dfrac{\vec{n}}{|\vec{n}|}$$
A: The problem does not ask you to find a tangent plane! It asks you to find points where the normal vector is parallel to <3, 2, 2>.  The normal vector at any point of f(x,y,z)= constant is $\nabla f$.  Here $f(x, y, z)= 3x^2- 4y^2- z= 0$.  Find $\nabla f$ and set it equal to  <3k, 2k, 2k> for some k. 
A: The other answers already covered the basics: you don’t need to find any tangent planes per se, but only points at which the normal to the surface is parallel to $n$. Since you’re working in $\mathbb R^3$, you have a bit of a short cut available: two nonzero vectors are parallel iff their cross product vanishes. Thus, you can avoid introducing another variable by stating the condition in the problem as $\nabla F\times n=0$, where $F:(x,y,z)\mapsto 3x^2-4y^2-z$. This generates three equations (only two of which are independent) to solve together with the original implicit equation of the surface.
