Let $X$ be an algebraic complex K3 surface, we know that $X$ is deformation equivalent to a smooth quartic surface or more generally a K3 surface with Picard number $1$ (a very general K3 surface in the family of K3 surfaces). It means that there exists a proper morphism $\mathcal{X}\rightarrow S$ with each fibre a K3 surface, $\mathcal{X}\_{\eta}=X$ ($\eta$ a general point on $S$) and $\rho(\mathcal{X}\_t)=1$ for some closed $t\in S$.



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> I want to show that $S$ can be a smooth algebraic curve or better be $\mathbb{A}^1$.
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One-dimensional family of complex algebraic K3 surfaces
Choose any ample class in $\mathrm{Pic}(X)$, assume its degree is $d$. Let $M\_d$ be the moduli space of polarized K3 surfaces of degree $d$ with appropriate level structure so that it has a universal family (alternatively one can use here the $\mathrm{Quot}$-scheme that is used to construct the moduli space). Let $x \in M\_d$ be the point that corresponds to $X$. Let $S\_0 \subset M\_d$ be a general curve in $M\_d$ through $x$, and let $S \to S\_0$ be a resolution of singularities of $S\_0$. Then the pullback to $S$ of the universal family from $M\_d$ is a family with the required properties.


