Angular momentum about a translating point carries a transport term

In an inertial frame let a reference point have position $\mathbf R(t)$ and velocity $\mathbf V(t)$. Define $\mathbf L_R=\sum_i(\mathbf r_i-\mathbf R)\times m_i\mathbf v_i$ and $\boldsymbol\tau_R=\sum_i(\mathbf r_i-\mathbf R)\times\mathbf F_i$. Derive the exact balance for $\dot{\mathbf L}_R$ without assuming that $\mathbf R$ is fixed. For $\boldsymbol\tau_R=(0,0,12.0)\,\mathrm{N\,m}$, $\mathbf V=(3.00,0,0)\,\mathrm{m/s}$, and total momentum $\mathbf P=(0,4.00,0)\,\mathrm{kg\,m/s}$, compute $\dot L_{R,z}$. State the special choice of $\mathbf V$ that removes the transport term.

Differentiate each relative position $\mathbf r_i-\mathbf R$ explicitly.