In the projective plane of any field , any cubic curve intersects a straight line at exactly $3$ points?

Question

Let $k$ be a field $f\in k[x,y,z]$ be a homogenous polynomial of degree $3$, let $C:=\{(x:y:z)\in P(2,k):f(x,y,z)=0\}$ , where $P(2,k)$ denotes the projective plane . Let $a,b,c\in k$ and $E:=\{(x:y:z)\in P(2,k): ax+by+cz=0\}$ , then is it true that $|C\cap E|=3$ ?

Answer 1

No. For instance, if $f(x, y, z) = yz^2 - x^3$ and $E = \{(x: y: z)\mid y = 0\}$, then there is only one intersection point, namely $(0:0:1)$. The intersection has multiplicity $3$, though.

In general, when counted with multiplicity, if $f(x, y, z)$ is homogenuous of degree $m$ and $g(x, y, z)$ is homogenuous of degree $n$, $f$ and $g$ have no common factor, and $k$ is algebraically closed, the zero sets of $f$ and of $g$ intersect $m\cdot n$ times in $P(2, k)$. This comes from a theorem called Bézout's theorem.