Suppose $x,y,w,z$ are homogeneous coordinates of $\mathbb{CP}^3$, and \begin{eqnarray} X_t := \left(F_t = x f_2 +y g_2 +t F_3 = 0 \right) \end{eqnarray}

be a family of degree 3 hypersurfaces in $\mathbb{CP}^3$, and $f_2, g_2$ and $F_3$ are generic homogeneous polynomials of degree 2,2 and 3.

What I often read in physics literature is that on $t=0$, the line $x=y=0$ lies entirely inside $X_0$, and we have a ``new" algebraic $(1,1)$ cycle, and if we deform $X_0$ to some $X_t$ (with small $|t|$), that cycle transform into a mixed cycle.

I want to understand that, i.e. I want to know

1-Why $x=y=0$ is a new $(1,1)$ cycle. Usually, I read this is because this cycle isn't a complete intersection of a hyperplane in $\mathbb{CP}^3$ with $X_0$. But I'm not sure whether this is enough to say that this algebraic cycle is independent of the other cycles.

2- Why when I (infinitesimally) deform $X_0$ to $X_t$, the cycle transforms into a mixed one. Again in physics, people would say, well $x=y=0$ is just three points inside $X_t$. But it would be nicer to see this more in a more ``intrinsic" way.

I tried to find the answers by using the Picard-Fuchs equation, but no success yet.

For example, we have the correspondence (the following theorem is due to Griffiths if I'm right),

\begin{eqnarray} Prim^{1,1}(X_t) = \left(\frac{\mathbb{C}[x,y,z,w]}{<\frac{\partial F_t}{\partial x},\frac{\partial F_t}{\partial y},\frac{\partial F_t}{\partial z},\frac{\partial F_t}{\partial w}>} \right)_{2}, \end{eqnarray} i.e. the prime elements of $H^{1,1}(X_t)$ correspond to homogeneous polynomials of degree 2 of the right hand side in the above equation.

In this case, we find $dim_{\mathbb{C}} Prim^{1,1}(X_t) = 6$, which I think it's true since $h^{1,1}(X_t) = 7$, so primimitivity must put one contraint and we get 6 independent prime lements.

Now, I expect that $dim_{\mathbb{C}} Prim^{1,1}(X_0)$ to jump (because I must have extra cycles), but it stays the same!