3. Given that 10 prime numbers form an arithmetic sequence. Then their common difference $d$ must satisfy ( ).
(A) $d$ is a multiple of 2, 3, 5, 7
(B) $d$ is a multiple of 11
(C) $d \leqslant 200$
(D) $d \leqslant 50$
3. $(\mathrm{A})$.

Obviously, 2 is not a number in this sequence. Next, we prove that $3 \mid d$. In $a, a+d, a+2d (a, d \in \mathbf{N})$, if $3 \nmid d$, then the remainders when these three numbers are divided by 3 must be different, otherwise $3 \mid d$. In this case, $a, a+d, a+2d$ must include a multiple of 3, which contradicts that they are prime numbers. Similarly, we can prove that $5 \mid d, 7 \mid d$, but we cannot conclude that $11 \mid d$.