Can I assume that L1 is regular? $L_{4},L_{2}$ are regular languages.

given an expression which we know is regular: $L_{4}\cap \bar{L_{2}}\cap \bar{L_{1}}$

May I assume that the language $L_1$ is regular? using closure properties
No, in general. If $\bar{L_2}$ and $L_4$ are both the empty language, or have an empty intersection, then $L_4\cap\bar{L_2}\cap\bar{L_1}$ is regular, but that says nothing about whether $L_1$ is regular.