Let $V_i\subseteq \mathbb{P}^d$ be an irreducible algebraic variety of dimension $d_i$ and degree $n_i$. Bezout's theorem says that if $d_1+d_2=d$ then $$V_1\cap V_2$$ has $$n_1n_2$$ many points (counting multiplicity.)
${\bf Question\ 1:}$ Is it true that if $d_1+d_2+d_3=d$ then
$$V_1\cap V_2 \cap V_3$$ has
$$n_1n_2n_3$$ many points (counting multiplicity.) ?
Have found an old book that affirms this.
Here is an example that refutes it.
${\bf Example:}$ Let the points of $\mathbb{P}^5$ be identified as the coefficients of a projective conic in the plane. Let $P$ be a hyperplane in $\mathbb{P}^5$ consisting of those conics passing through a given point. The statement that a conic is determined by $5$ points I write as $$P^5=1$$ meaning that if I intersect five of these hyperplanes, defined by distinct points of course, but still all denoted by $P$, I get a single point. The intersection of $5$ hyperplanes in $5$ space being a point (in general).
Now let $L$ be the surface of conics tangent to a given line. It is known $$P^4L=2$$ meaning that there are two conics through $4$ given points and tangent to a given line. One sees that this implies that $L$ is a quadratic hypersurface. It follows from Bezout that $$P^3L^2=4$$ and this is ${\it correct}$. It then follows by duality that $$P^2L^3=4$$ $$PL^4=2$$ $$L^5=1$$ and these values are ${\it correct}$.
Jacob Steiner however at one point made the mistake that $$P^2L^3=2^3$$ $$PL^4=2^4$$ $$L^5=2^5$$
But doesnt this follow from the generalized Bezout theorem ?
${\bf Question 2:}$ What is the error is Steiner's supposition ?