There are many questions like this, but none of them expose my error.
Given the surface: $$\frac{x^2}{9}+\frac{y^2}{256}+\frac{z^2}{2304}=1$$
Find the points where the tangent plane is parallel to the plane $2x-3y+z=5$.
I found the gradient of the surface: $\left\langle \frac {2x}{9}, \frac{2y}{256}, \frac{2z}{2304} \right\rangle$
I found the normal to the plane: $\langle 2,-3,1 \rangle$
I solved for $(a,b,c)$ with:
$$\left\langle \frac {2x}{9}, \frac{2y}{256}, \frac{2z}{2304} \right\rangle=\langle 2,-3,1 \rangle$$
I determined the point to be $(9,-384,1152)$. This is apparently wrong, though I see no arithmetical or algebraic errors. I also tried the antiparallel point $(-9,384,-1152)$, which does not even seem like a correct notion. Are there more points?
I would appreciate any help as to what corrections I need to make.
Thank you to @WillJagy for answering in the comments. I am answering this to close the question.
Let $\left\langle \frac {2x}{9}, \frac{2y}{256}, \frac{2z}{2304} \right\rangle= \lambda ⟨2,−3,1⟩$
Instead of a point $(a,b,c)$, this results in $P: (\lambda a,\lambda b,\lambda c)$.
Plug $P$ back into the surface $\frac{x^2}{9}+\frac{y^2}{256}+\frac{z^2}{2304}=1$ and solve for the value(s) of $\lambda$. Plug the resulting value(s) for $\lambda$ back into $P: (\lambda a,\lambda b,\lambda c)$ for the points.