1. Achievability
We note the lower bound of storage constrained PIR with $\mu$storage coefficient. We have:
\[\begin{align} D\geq&L+\dfrac{1}{2}\sum_{k=1}^{2}\sum_{\mathcal{S}:|\mathcal{S}|= 1}|W_{k,\mathcal{S}}|L+\dfrac{5}{12}\sum_{k=1}^{2}\sum_{\mathcal{S}:|\mathcal{S}|= 2}|W_{k,\mathcal{S}}|L+\dfrac{1}{3}\sum_{k=1}^{2}\sum_{\mathcal{S}:|\mathcal{S}|= 3}|W_{k,\mathcal{S}}|L\\ =&L+\sum_{\mathcal{S}:|\mathcal{S}|= 1}\alpha_{\mathcal{S}}+\dfrac{5}{6}\sum_{\mathcal{S}:|\mathcal{S}|= 2}\alpha_{\mathcal{S}}+\dfrac{2}{3}\sum_{\mathcal{S}:|\mathcal{S}|= 3}\alpha_{\mathcal{S}}\\ =&2\sum_{\mathcal{S}:|\mathcal{S}|= 1}\alpha_{\mathcal{S}}+\dfrac{11}{6}\sum_{\mathcal{S}:|\mathcal{S}|= 2}\alpha_{\mathcal{S}}+\dfrac{5}{3}\sum_{\mathcal{S}:|\mathcal{S}|= 3}\alpha_{\mathcal{S}} \end{align}\]which is equivalent to the following linear problem:
\[\begin{align} \min_{\alpha_{\mathcal{S}}\geq 0} \quad &2(\alpha_1+\alpha_2+\alpha_3)+\dfrac{11}{6}(\alpha_{1,2}+\alpha_{2,3}+\alpha_{1,3})+\dfrac{5}{3}\alpha_{1,2,3} \\ s.t. \quad & \alpha_1+\alpha_2+\alpha_3+\alpha_{1,2}+\alpha_{2,3}+\alpha_{1,3}+\alpha_{1,2,3}=1 \\ &\alpha_1+\alpha_{1,2}+\alpha_{1,3}+\alpha_{1,2,3} \leq \mu\\ &\alpha_2+\alpha_{1,2}+\alpha_{2,3}+\alpha_{1,2,3} \leq \mu\\ &\alpha_3+\alpha_{2,3}+\alpha_{1,3}+\alpha_{1,2,3} \leq \mu\\ \end{align}\]the fractions are all easy according to TPIR scheme and memory sharing scheme. Every $\alpha_{1,2,3},\alpha_{1,2},\alpha_{1}$sums are all achievable values. In fact we can figure out the coefficients in the following CONVERSE result:
\[\begin{equation} x\sum_{k=1}^{2}H(W_k)+y\sum_{k=1}^{2}\sum_{i\neq k}H(W_k|Z_i)+z\sum_{k=1}^{2}H(W_k|Z_{[3]-\{k\}}) \end{equation}\]via
\[\begin{align} &2x+4y+2z=2,\quad |\mathcal{S}|=1 \\ &2x+2y=\dfrac{11}{6},\quad |\mathcal{S}|=2 \\ &2x=\dfrac{1}{3},\quad |\mathcal{S}|=3 \end{align}\]and $(x,y,z)=(\dfrac{5}{6},\dfrac{1}{12},0)$
2.Converse
\[\begin{align} H(A_{[3]}^1|W_1)=&H(A_1^1,A_2^1|W_1)+H(A_3^1|A_1^1,A_2^1,W_1) \end{align}\]by taking all possible permutation $(i,j,k)\sim(1,2,3)$together,
\[\begin{align} 6H(A_{[3]}^1|W_1)=&2(H(A_1^1,A_2^1|W_1)+H(A_1^1,A_3^1|W_1)+H(A_3^1,A_2^1|W_1))\\ &+2(H(A_3^1|A_1^1,A_2^1,W_1)+H(A_2^1|A_1^1,A_3^1,W_1)+H(A_1^1|A_2^1,A_3^1,W_1)) \end{align}\]However, according to $(11)\sim(13)$, the ideal coefficient
\[\begin{align} D \geq &\dfrac{5}{6}\sum_{k=1}^{2}H(W_k)+\dfrac{1}{12}\sum_{k=1}^{2}\sum_{i\neq k}H(W_k|Z_i)\\ =&L+\dfrac{1}{3}\sum_{k=1}^{2}H(W_k)+\dfrac{1}{12}\sum_{k=1}^{2}\sum_{i\neq k}H(W_k|Z_i) \end{align}\]which is equivalent to
\[\begin{align} 6H(A_{[3]}^1|W_1) \geq &4H(W_2)+(H(W_2|Z_1)+H(W_2|Z_2)+H(W_2|Z_3)) \\ =&2(H(A_1^1,A_2^1|W_1)+H(A_1^1,A_3^1|W_1)+H(A_3^1,A_2^1|W_1))\\ &+(H(A_1^1,A_2^1|W_1,Z_3)+H(A_1^1,A_3^1|W_1,Z_2)+H(A_3^1,A_2^1|W_1,Z_1)) \end{align}\]and
\[\begin{align} &H(A_3^1|A_1^1,A_2^1,W_1)+H(A_1^1|A_2^1,A_3^1,W_1)\geq H(A_1^1,A_3^1|W_1,Z_2)=H(A_3^1|A_1^1,Z_2,W_1)+H(A_1^1|Z_2,W_1)\\ \Leftrightarrow & I(A_1^1;A_3^1|Z_2,W_1)=0 \end{align}\]