Let $\hat M(\vec x)=\sum_{i=1}^n x_k\hat M_k$ be an $n\times n$ matrix pencil such that $\det M(\vec x)=0$ for every $\vec x$. The $n$ matrices $\hat M_k$ are linearly independent, that is, $\vec M(\vec x)$ is not equal to zero for every $\vec x$. Furthermore, we have that $\hat M(\vec x)\vec y=\hat M(\vec y)\vec x$ for every $\vec x$ and $\vec y$. I want to prove that there is a vector $\vec z\ne\vec 0$ such that $\vec z^t \hat M(\vec x)=0$ for every $\vec x$.
If this conjecture is correct, it would be also nice to weaken the hypotheses.
This is false. I think this is better discussed using tensors or trilinear forms, not matrices, so let me change the language.
Denote $F(\vec{x}, \vec{y}, \vec{z}) = \vec{z}^t \hat{M}(\vec{x}) \vec{y}$. Clearly, $F$ is linear in each of the three vectors. We can go back to $\hat{M}(\vec{x}) = (m_{ij}(\vec{x}))$ by setting $m_{ij}(\vec{x}) = F(\vec{x}, \vec{e}_i, \vec{e}_j)$, where $\vec{e}_i$ are basis vectors.
Now the symmetry condition becomes $F(x, y, -) = F(y, x, -)$, $\det \hat{M}(x) = 0$ means that for each $x$ the rank of the bilinear form $F(x, -, -)$ is less than $n$, and we want to show that $F(-, -, z) = 0$ for some $z \neq 0$.
We only need symmetry in $x, y$, but the counterexample will actually be fully symmetric in all three variables. Take $n = 5$ and $$\begin{aligned}F(x, y, z) =\, & x_0 y_0 z_2 + (x_0 y_1 + x_1 y_0) z_3 + x_1 y_1 z_4 \\ +\, & x_0 z_0 y_2 + (x_0 z_1 + x_1 z_0) y_3 + x_1 z_1 y_4 \\ +\, & y_0 z_0 x_2 + (y_0 z_1 + y_1 z_0) x_3 + y_1 z_1 x_4 \\ \end{aligned}$$ This example is adapted from a classical example of a cubic form with vanishing hessian determinant.
The corresponding matrix $\begin{bmatrix} x_2 & x_3 & x_0 & x_1 & 0 \\ x_3 & x_4 & 0 & x_0 & x_1 \\ x_0 & 0 & 0 & 0 & 0\\ x_1 & x_0 & 0 & 0 & 0\\ 0 & x_1 & 0 & 0 & 0 \end{bmatrix}$ clearly has rank $4$ and does not vanish for any nonzero $x$, so $F(-,-,z) = F(z,-,-)$ is also nonzero for every nonzero $z$. We have a counterexample.