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L1 is regular,L is regular then $ L1\cap L2$  is also a regular because regular is closed under intersection.

https://www.geeksforgeeks.org/closure-properties-of-regular-languages/

 it is also closed under union, concatenation, kleen closure, set difference, positive closure, complement, reverse operator, homomorphism, inverse homomorphism etc

if L1 and L2 are regular languages, then each of also $L1\cup L2$ , $L1.L2$ and $L1^{*}$ is regular etc.

option A

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Option A) is true as intersection is closed in regular grammars.
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