Can this matrix be negative definite?

Let $$d = 12$$ and $$m = 6$$, and denote by $$0_n$$ and $$I_n$$ the zero matrix and the identity matrix of size $$n \times n$$.

Let $$D_+ \in \mathbb{R}^{m \times m}$$ be a diagonal matrix with positive diagonal entries and $$D \in \mathbb{R}^{d \times d}$$ be defined by \begin{align*} D = \begin{bmatrix} -D_+ & 0_m\\ 0_m & D_+ \end{bmatrix} \end{align*} Let $$B = B(x) \in \mathbb{R}^{d \times d}$$ have the form \begin{align*} B = \frac{1}{2} (D \bar{P}+ \bar{J}^{-1}\bar{P}^\intercal\bar{J}D), \qquad \text{with }\bar{P} = \begin{bmatrix} P & P \\ P & P \end{bmatrix}, \quad \bar{J} = \begin{bmatrix} \hat{J} & 0_m \\ 0_m & \hat{J} \end{bmatrix}, \end{align*} where $$P,J \in \mathbb{R}^{m \times m}$$ are defined by \begin{align*} P = \begin{bmatrix} \Upsilon & \Gamma \\ 0_3 & \Upsilon \end{bmatrix}, \quad \hat{J} = \begin{bmatrix} I_3 & 0_3 \\ 0_3 & J \end{bmatrix} \end{align*} for $$\Upsilon, \Gamma \in \mathbb{R}^{3 \times 3}$$ skew-symmetric matrices, $$J \in \mathbb{R}^{3 \times 3}$$ a diagonal matrix with positive diagonal entries.

My question is:

Is it possible to find a diagonal matrix $$Q = Q(x) \in \mathbb{R}^{d \times d}$$ (for $$x \in [0,L]$$), such that the matrix $$M = M(x)\in \mathbb{R}^{d \times d}$$ \begin{align*} M(x) = Q'(x)D + Q(x)B + B^\intercal Q(x) \end{align*} is negative definite for any $$x \in [0,L]$$ ?

(Here $$B^\intercal$$ denotes the transpose of $$B$$.)

Remark: the diagonal of $$B$$ contains only zeros (since $$\Upsilon$$ is skew-symmetric), hence the diagonal of $$QB+B^\intercal Q$$ contains only zeros.

Any suggestion or reference is welcome. Thank you.