problem
Find all points on the surface $$ f(x,y,z) = x^2 + 2y^2 + 3z^2 + 2xy + 2yz = 1 $$ where the tangent plane is parallel to the plane $$ x - y + 2z = 0 $$
NOTE THIS EDIT
I had made a typo ($3yz$ instead of $2yz$) in the original equation. I will try solving it again now and then see what I will do about this question since the rest from here might be irrelevant (except for the answer).
my try at solving it
The normal vector $n_1$ of the plane is $(1,-1,2)$ and I compute the gradient of the surface to the following (square brackets for emphasizing components) $$ \nabla f(a,b,c) = ([2(a+b)], [2a+4b+3c], [3b+6c]) $$ which is a normal vector $n_2$ of the point $(a,b,c)$ on the given surface.
Then, like all other find-all-points-questions I try to compute the cross product $n_2 \times n_1$ since it should equal to the zero-vector if they are parallel. From this, however, I get the set of equations $$ \begin{cases} 4a + 11b + 12c = 0\\ 4a + b - 6c = 0\\ 3b + 6c = 0 \end{cases} $$ which I see doesn't lead to anything but $a=b=c=0$ since there are no constants involved (incorrect if I'm not missing something here).
my other try
I also tried to set the $n_2$ components equal to the $n_1$ components, since they should point in the same direction, and then I think I can scale the expression by inserting the points in the original function $f$ and scale them to equal $1$. So I set up the following system of equations $$ \begin{cases} 2a + 2b = 1\\ 2a + 4b + 3c = -1\\ 3b + 6c = 2 \end{cases} $$ and solve it using Gauss-Jordan giving me $$ \begin{cases} a=13/2\\ b=-6\\ c=10/3 \end{cases} $$ and I get the tedious expression $$ 13^2/4 + 72 + 100/3 + 13(-6) - 6\cdot 10 = 115/12 $$ which also is incorrect, since the answer is supposed to be $$ \pm \left( \frac{5}{\sqrt{13}}, \frac{-4}{\sqrt{13}} ,\frac{2}{\sqrt{13}} \right) $$ which I can't get from scaling the points $a,b,c$ with the factor $115/12$. I have gone through this exercise several times, I have compared with similar exercises and I don't know what to do. Can anyone see what I'm missing or point at where I'm doing wrong? Is my approach even correct for this specific question?
should be $$ \begin{cases} 2a + 2b = \lambda\\ 2a + 4b + 3c = -\lambda\\ 3b + 6c = 2 \lambda \end{cases} $$ $$ \begin{cases} a=13 \lambda/2\\ b=-6 \lambda\\ c=10 \lambda/3 \end{cases} $$ or back to $$ \begin{cases} x=13 \lambda/2\\ y=-6 \lambda\\ z=10 \lambda/3 \end{cases} $$ with $$x^2 + 2y^2 + 3z^2 + 2xy + 3yz = 1$$ Fewer errors if we define $$ \lambda = 6t $$ and $$ \begin{cases} x=39t\\ y=-36 t\\ z=20t \end{cases} $$ $$x^2 + 2y^2 + 3z^2 + 2xy + 3yz = 345 t^2$$ and $$ t = \pm \frac{1}{ \sqrt {345} } $$