Let $S$ be the set of all numbers which are the sum of the squares of three consecutive integers. Then we can say that
$\text{(A) No member of S is divisible by } 2\quad\\ \text{(B) No member of S is divisible by } 3 \text{ but some member is divisible by } 11\quad\\ \text{(C) No member of S is divisible by } 3 \text{ or } 5\quad\\ \text{(D) No member of S is divisible by } 3 \text{ or } 7 \quad\\ \text{(E) None of these}$
Consider $3$ consecutive integers $a, b,$ and $c$.  Exactly one of these integers must be divisible by 3; WLOG, suppose $a$ is divisible by 3.  Then $a \equiv 0 \pmod {3},  b \equiv 1 \pmod{3},$ and $c \equiv 2 \pmod{3}$.  Squaring, we have that $a^{2} \equiv 0 \pmod{3}, b^{2} \equiv 1 \pmod{3},$ and $c^{2} \equiv 1 \pmod{3}$, so $a^{2} + b^{2} + c^{2} \equiv 2 \pmod{3}$.  Therefore, no member of $S$ is divisible by 3.
Now consider $3$ more consecutive integers $a, b,$ and $c$, which we will consider mod 11.  We will assign $k$ such that $a \equiv k \pmod{11}, b \equiv k + 1 \pmod{11},$ and $c \equiv k + 2 \pmod{11}$.  Some experimentation shows that when $k = 4, a \equiv 4 \pmod{11}$ so $a^{2} \equiv 5 \pmod{11}$.  Similarly, $b \equiv 5 \pmod{11}$ so $b^{2} \equiv 3 \pmod{11}$, and $c \equiv 6 \pmod{11}$ so $c^{2} \equiv 3 \pmod{11}$.  Therefore, $a^{2} + b^{2} + c^{2} \equiv 0 \pmod{11}$, so there is at least one member of $S$ which is divisible by 11.  Thus, $\fbox{B}$ is correct.