Which of the following languages are undecidable? Note that $\langle M \rangle$ indicates encoding of the Turing machine M. $L_1 = \{\langle M \rangle \mid L(M) = \emptyset\}$ $L_2 = \{\langle M, w, q \rangle \mid M \text{ on input } w \text{ reaches state } q \text{ in exactly 100 steps}\}$ $L_3 = \{\langle M \rangle \mid L(M) \text{ is not recursive}\}$ $L_4 = \{\langle M \rangle \mid L(M) \text{ contains at least 21 members}\}$

To determine the decidability of each language:

1. Language $L_1 = \{\langle M \rangle \mid L(M) = \emptyset\}$: This is the emptiness problem for Turing machines. By Rice's theorem, any non-trivial semantic property of the language recognized by a Turing machine is undecidable. Emptiness is non-trivial, so $L_1$ is undecidable.

2. Language $L_2 = \{\langle M, w, q \rangle \mid M \text{ on input } w \text{ reaches state } q \text{ in exactly 100 steps}\}$: Since the number of steps is strictly bounded by 100, we can simulate $M$ on input $w$ for 100 steps and check whether state $q$ is reached at the $100^{\text{th}}$ step. This simulation always halts in finite time, making $L_2$ decidable.

3. Language $L_3 = \{\langle M \rangle \mid L(M) \text{ is not recursive}\}$: Testing whether the language of a Turing machine is recursive is a non-trivial property of recursively enumerable languages. By Rice's theorem, $L_3$ is undecidable.

4. Language $L_4 = \{\langle M \rangle \mid L(M) \text{ contains at least 21 members}\}$: Cardinality of the recognized language is a non-trivial semantic property. By Rice's theorem, $L_4$ is undecidable.

Therefore, the undecidable languages are $L_1, L_3,$ and $L_4$ only, making A the correct option.