In $\mathbb{R}^n$, given an unit $L_1$ sphere $\mathcal{B}_n: x_1+x_2+\ldots+x_n\leq 1$ and a hyperplane $\mathcal{P}: a_1x_1+a_2x_2+\ldots+a_nx_n=0$. Does there always exist a rotation such that $\mathcal{B}_n\cap\mathcal{P}$ is embedded into the $(n1)$ dimensional unit $L_1$ sphere: $\mathcal{B}_{n1}: x_1+x_2+\ldots+x_{n1}\leq 1, x_n=0$ after the rotation?

Thanks for the comments above. I have just proofed this problem is only true when $n\leq 4$. When $n\leq 4$, without loss of generality, assume $a_n\geq a_1, \ldots, a_{n1}$. Let $Q$ be the rotation on the plane spanned by $(a_1, \ldots, a_n)$ and $(0,\ldots, 0, 1)$ such that applying $Q$ to $(a_1, \ldots, a_n)$ rotates it to $(0,\ldots, 0, 1)$. Apply the rotation to $\mathcal{B}\cap\mathcal{P}$ will inscribe it into $\mathcal{B}_{n1}$. For $n=5$, the intersection between $\mathcal{B}_n$ and $\mathcal{P}: a_1+a_2+a_3+a_4+a_5=0$ can not be inscribed in to the 4dimension $L_1$ sphere. The intuition of the proof is that the $\mathcal{B}_n\cap\mathcal{P}$ contains the intersection between the 3dimension $L_1$ sphere and hyperplane $a_1+a_2+a_3+a_4=0$. This intersection can only be inscribed into the 4dimensional $L_1$ sphere under two rotations (with symmetry cases ignored). One can verified that non of those two rotations can fit $\mathcal{B}_n\cap\mathcal{P}$ into a 4dimensional $L_1$ sphere. 

