As $\vec{F}=q(\vec{v} \times \vec{B})$ - $$ \vec{a}=\frac{q}{m}(\vec{v} \times \vec{B}) $$ so, $\overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{B}}$ are ⟂ to each other Hence, $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{B}}=0$ $$ \begin{aligned} & (\alpha \hat{\mathrm{i}}-4 \hat{\mathrm{j}}) \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}})=0 \\ & \alpha(2)+(-4)(3)=0 \\ & \alpha=\frac{12}{2} \Rightarrow \alpha=6 \end{aligned} $$