Let $\sum_{i=1}^n x_k \hat M_k\equiv \hat M(\vec x)$ be a $n\times n$ matrix pencil such that $\det \hat M(\vec x)=0$ for every $\vec x$. Let us also assume that for every vector $\vec y$, there is a vector $\vec x$ such that $\hat M(\vec x)\vec y=0$. An obvious way for satisfying these conditions is to choose general matrices for $\hat M_k$ such that their images are equal to a given $(n-1)$-dimensional vector space. In other words, this means that there is a vector $\vec z\ne\vec 0$ such that $\hat M_k^T \vec z=0$ for every $k=1,\dots,n$. Is this constraint necessary?
A weaker condition is that there is a basis $\vec y_1,\dots,\vec y_n$ such that there are vectors $\vec x_k$ with $\hat M(\vec x_k)\vec y_k=0$ for every $k=1\dots n$.
No. Consider the pencil $\begin{bmatrix}0 & x_1 & x_2 \\ -x_1 & 0 & x_3 \\ -x_2 & -x_3 & 0\end{bmatrix}$ generating all alternating $3 \times 3$ matrices.
The determinant is $0$ (every alternating matrix has rank $2$). For every $\vec{y} = (y_1, y_2, y_3)^\top$ it is easy to find a matrix $\hat{M}(\vec{x})$ with $\vec{y}$ in the kernel, namely $\begin{bmatrix}0 & y_3 & - y_2 \\ - y_3 & 0 & y_1 \\ y_2 & -y_1 & 0\end{bmatrix}$ for $\vec{y} \neq 0$.
But the common kernel of all alternating matrices is trivial, so we have a counterexample.