This question did not get up votes nor answers, maybe because as it turns out, this was not that complicated. In any case, I still propose a detailed answer just in case.
Solving for $\dot x$ and $\dot \theta$
The impulse applied on top would be defined as the variation in the linear momentum of the particles system. For more convenience, we will be using particle $B$ as a reference point.
\begin{equation} \hat{\mathbf{J}} = \sum\limits_{i=1}^{3}\Delta \mathbf{p_i} \end{equation}
where $\hat{\mathbf{J}}$ is the total impulse vector affecting the system, comprised of the horizontal $\hat{F}$ and the vertical constraint impulse $\hat{N}$; and $\sum\limits_{i=1}^{3}\Delta \mathbf{p_i}$ is the sum of the linear momentum variation of the system particles. 804
Since the system is initially at rest, its particles initial velocities are null, so the linear momentum variation is simply the sum of their individual final momenta.
Because particles $A$ and $C$ are subject to a linear velocity $\dot x$ resulting from the sliding of particle $B$; and to an angular velocity $\dot\theta$ resulting from the rotation about particle $B$; the above equation can then be developed as:
\begin{equation} \left(\hat{F}\,\vec{i}+\hat{N}\,\vec{j}\right) = m\left( \left( -l\frac{\sqrt{3}}{2}\vec{i} + \frac{l}{2}\vec{j} \right) \times \dot\theta\vec{k} + \dot x\vec{i} \right) + m\left( l\vec{j} \times \dot\theta\vec{k} + \dot x\vec{i} \right) + m\dot x\vec{i} \end{equation}
where $\vec{i}$, $\vec{j}$ and $\vec{k}$ are unit vectors of the frame and the $\times$ operator denotes the cross product between vectors. On the right hand side of the above equation, the first term corresponds to the linear momentum of particle $C$, the middle one corresponds to the linear momentum of particle $A$ and the last term is the linear momentum of particle $B$ which is only subject to a sliding motion.
By developing the above equation, the $\hat{F}$ and $\hat{N}$ components of the applied linear impulse vector $\hat{\mathbf{J}}$ are identified as the horizontal linear impulse applied on particle $A$:
\begin{equation} \hat{F} = 3m\dot x + \frac{3}{2}m l \dot\theta \end{equation}
and the constraint linear impulse applied by the supporting base on particle $B$:
\begin{equation} \hat{N} = \frac{\sqrt{3}}{2}m l \dot\theta \end{equation}
Since the constraint linear impulse $\hat{N}$ is considered a consequence of the linear impulse $\hat{F}$ and the value of $\hat{N}$ is unknown, we still need one equation to solve for the $\dot x$ and $\dot \theta$ values. Keeping particle $B$ as the reference point for the angular momentum of each particle, we can find $\dot x$ and $\dot \theta$ without having to solve for $\hat{N}$. Thus, we can state a relationship between the angular impulse at $B$ resulting from the linear impulse on particle $A$ and the variation of the angular momentum of each particle:
\begin{equation} \hat{\mathbf{M}} = \sum\limits_{i=1}^{3}\Delta \mathbf{H_i} = \sum\limits_{i=1}^{3}\mathbf{r}_i \times \Delta \mathbf{p_i} \end{equation}
where $\hat{\mathbf{M}}$ stands for the angular impulse around particle $B$ of the linear impulse $\hat{F}$ applied on $A$; $\sum\limits_{i=1}^{3}\Delta \mathbf{H_i}$ represents the sum of the individual angular momenta around $B$ of particles $A$ and $C$; and $\mathbf{r}_i$ is the vector going from $B$ to the other particles. 804