Let $C$, $Q \in \mathbb{R}[x_0,\dots,x_n]$ be homogeneous of degrees $3$ and $2$ respectively. Consider the scheme $V$ in $\mathbb{P}^n$ defined by $$ V \; : \; C=Q=0$$. Suppose
Question: Show that $V$ has a smooth real point.
Note: It is easy to construct a counterexamples without the integrality assumption. For example, take $C=x_0^3+x_1 Q$. Then every point on $V$ has multiplicity $3$.
I think that this is wrong for $n = 3$:
Let $C \subset {\mathbb P}^3_{\mathbb R}$ be a generic cubic surface not containing $(0:0:0:1)$ and take $$Q \colon x^2 + y^2 + z^2 = \varepsilon w^2$$ with $\varepsilon > 0$ sufficiently small (the coordinates are $x,y,z,w$). Then $C({\mathbb R}) \cap Q({\mathbb R})$ is empty, but for most choices of $C$ (and $\varepsilon$) $V = C \cap Q$ will be integral over $\mathbb C$.
For larger $n$, the same kind of example works.
If the signature of $Q$ is $(2,2)$ instead of $(3,1)$, then the statement is true: $Q$ is isomorphic over $\mathbb R$ to ${\mathbb P}^1 \times {\mathbb P}^1$ and $V$ gets mapped to a curve of type $(3,3)$, which will always have lots of real points. So perhaps one needs an additional assumption on the signature of $Q$.
