6. Let $\left\{\left(s_{1}, s_{2}, \cdots, s_{6}\right) \mid s_{i} \in\{0,1\}, i \in \mathbf{N}_{+}, i \leqslant 6\right\}$, for $\forall x, y \in S$, $x=\left(x_{1}, x_{2}, \cdots, x_{6}\right), y=\left(y_{1}, y_{2}, \cdots, y_{6}\right)$, define:
(1) $x=y$ if and only if $\left(x_{1}-y_{1}\right)^{2}+\left(x_{2}-y_{2}\right)^{2}+\cdots+\left(x_{6}-y_{6}\right)^{2}=0$;
(2) $x \cdot y=x_{1} y_{1}+x_{2} y_{2}+\cdots+x_{6} y_{6}$.

If a non-empty set $T \subseteq S$, and satisfies $\forall u, v \in T, u \neq v$, then $u \cdot v \neq 0$, the maximum number of elements in the set $T$ is
A. 62
B. 52
C. 42
D. 32
When $u, v$ contain at least 3 ones, the number of elements in $T$ is maximized. Among them, there is 1 element with 6 ones, 6 elements with 5 ones, 15 elements with 4 ones ($C_{6}^{2}=15$), and 20 elements with 3 ones ($C_{6}^{3}=20$). Clearly, elements with different numbers of ones have a product that is not equal to 0, so elements with 3 ones can only take half of them. Therefore, the maximum number of elements in $T$ is 32, so the answer is $D$.