I have some trouble to do this excercise:
Given $S \subset \mathbb{R}^3$ the implicit surface defined as $$ S=\left\{(x, y, z) \in \mathbb{R}^3 \left\lvert\, x^2+x z+y z+\frac{1}{2} z^2=1\right.\right\} . $$
Find all the points $p \in S$ such that the tangent plane to $S$ at $p$ is orthogonal to $(2,0,1)$.
Well, first of all, how a plane can be orthogonal to a point? Am I misunderstanding the problem? I know how to find a generic $ T_p S$ with the function $f(x,y,z)=x^2+x z+y z+\frac{1}{2} z^2$ and writing down its gradient: $$ T_p S = \{(x,y,z) \in \mathbb{R}^3: (2x_0+z_0,z_0,x_0+y_0+2z_0).((x,y,z)-(x_0,y_0,z_0)) \} $$ where $p=(x_0,y_0,z_0)$, but I don't know how to add the information of orthogonality. I also try writing some generic $v\in T_pS$ and trying to get some info using the fact that $ <v,(2,0,1)>=0$, but it's not working. Can someone give me a hint?
$S$ is defined by the level surface $$f(x,y,z)=1$$ where $f(x,y,z)=x^2+xz+yz+\frac{1}{2}z^2$. The gradient of $f(x,y,z)$ is orthogonal to the surface, so in order to answer your question is sufficient to determine all the points $p=(\overline x,\overline y,\overline z)\in S$ such that $\nabla f(p)$ defines the same line spanned by the vector $(2,0,1)$. It's easy to see that: $$\nabla f(x,y,z)=\left(2x+z,z,x+y+z\right)$$ $\nabla f(x,y,z)$ and $(2,0,1)$ generate the same $1$-dimensional vectorial subspace if and only if there exist a scalar $\alpha\in\mathbb R$ such that $\nabla f(x,y,z)=\alpha(2,0,1)$, i.e.: $$\begin{cases}2x+z=2\alpha\\z=0\\x+y+z=\alpha\end{cases}\:\Longrightarrow\:\begin{cases}x=\alpha\\y=0\\z=0\end{cases}$$ Thus the points $p$ must have the form $p=(\alpha,0,0)$ with $\alpha\in\mathbb R$. We have to impose the condition $p\in S$: $$p=(\alpha,0,0)\in S\:\Longleftrightarrow\:\alpha^2=1\:\Longleftrightarrow\:\alpha=\pm1$$ In conclusion the only points of $S$ such that the tangent plane in that point is orthogonal to the vector $(2,0,1)$ are: $$p_\pm=(\pm1,0,0)$$