Find a point on the surface z = 16 - 4x^2 - y^2 at which the tangent plane is normal to the line given by r(t) = < 3+4t, 2t, 2-t >. Then find the equation of the normal line to the surface at this point.

I assume the first thing is to find the point where the line crosses the surface, I'm failing at this. The rest is just plugging in tangent plane formulas.

I know a few of you out there have seen this stuff before...

If you take the d/dx and d/dy of the surface equation you can come up with tangent vectors; take the cross product of them to get the normal to the surface, and to determine where the normal is parallel to r(t) you solve where the cross product of the normal vector and r(t) = 0

okay yeah, i'm still stuck.

"If you take the d/dx and d/dy of the surface equation"

I did this.

Oh, the tangent vectors will be of the form [1, 0, dz/dx] and [0, 1, dz/dy]