Given $\mu$, $\lambda_1$, $\lambda_2$, $\gamma_1$, ..., $\gamma_{n}$, which are all scalars. Suppose $\mu$, $\lambda_1$, $\lambda_2$ $\in \left(0,1\right)$, $0\leqslant \lambda_1<\mu<\lambda_2 \leqslant 1$, $\gamma_j \in \left[0,\lambda_1\right]\cup \left[\lambda_2,1\right], \forall j$. We have unique $p_1$ and $p_2$ such that $p_i \in [0,1],\forall i$, $p_1+p_2=1$, and $\mu=p_1 \lambda_1+p_{2}\lambda_{2}$.
Define several sets as follows.
$Q=\left\{(q_1,...,q_{n})|q_j \in [0,1],\forall j,\sum_{j=1}^{n}q_j=1,\mu=q_1 \gamma_1+...+q_{n}\gamma_{n}\right\}$.
$R_i=\left\{(r_{i1},...,r_{in})|r_{ij} \in [0,1],\forall j,\sum_{j=1}^{n}r_{ij}=1,\lambda_i=r_{i1} \gamma_1+...+r_{in}\gamma_{n}\right\}$
$S=\left\{p_1(r_{11},...,r_{1n})+p_2(r_{21},...,r_{2n})|(r_{i1},...,r_{in})\in R_i\right\}$
The question is whether $Q=S$ is true? ($S \subseteq Q$ is straightforward)
Note $\gamma_1$,...,$\gamma_{n}$ are not necessarily affine independent, so both $Q$ and $R_i$ may contain more than one element.
The intuition of this question is when we write $\mu$ as a convex combination of $\left(\gamma_1, ...,\gamma_n\right)$, is it always feasible to replicate this sequentially by first writing $\mu$ as a convex combination of $\lambda_1$ and $\lambda_2$ and then write $\lambda_i$ as a convex combination of $\left(\gamma_1, ...,\gamma_n\right)$.
I really appreciate your help, thank you very much!