3. Let the set $M=\{1,2, \cdots, 10\}, T$ be a set of some binary subsets of $M$, satisfying that for any two different elements $\{a, b\}, \{x, y\}$ in $T$, we have
$11 \nmid(a x+b y)(a y+b x)$.
Find the maximum number of elements in $T$.
3. The maximum number of elements in $T$ is 25.

Given that the elements of set $M$ are all coprime with 11, for any $\{a, b\} \in T$, there exist unique $k$ and $l$ such that
$$
k a \equiv b(\bmod 11), l b \equiv a(\bmod 11).
$$

Define $f(\{a, b\})=\{k, l\}$ (when $k=l$, define $f(\{a, b\})=\{k\}$).
By definition, we have
$$
k l a \equiv a(\bmod 11) \Rightarrow k l \equiv 1(\bmod 11).
$$

Thus, for any $\{a, b\} \in T$, we have $f(\{a, b\}) \in \{\{2,6\},\{3,4\},\{5,9\},\{7,8\},\{10\}\}$.
Let $M_{1}=\{\{a, b\} \in M \mid f(\{a, b\})=\{2,6\}\}$,
$M_{2}=\{\{a, b\} \in M \mid f(\{a, b\})=\{3,4\}\}$,
$M_{3}=\{\{a, b\} \in M \mid f(\{a, b\})=\{5,9\}\}$,
$M_{4}=\{\{a, b\} \in M \mid f(\{a, b\})=\{7,8\}\}$,
$M_{5}=\{\{a, b\} \in M \mid f(\{a, b\})=\{10\}\}$.
Consider $\{a, k a\}(a=1,2, \cdots, 10)$ under modulo 11.
It is easy to verify that when $2 \leqslant k \leqslant 9$,
$k^{2} \neq 1(\bmod 11)$.
Thus, these 10 pairs are distinct.
Therefore, $M_{1}, M_{2}, M_{3}, M_{4}$ each have 10 elements.
And $M_{5}=\{\{a, b\} \in M \mid a+b=11\}$, which has 5 elements.

On one hand, for any $\{a, b\} \in M_{1}, \{x, y\} \in M_{3}$, assume
$2 a \equiv b(\bmod 11), 5 x \equiv y(\bmod 11)$.
Then $a x + b y \equiv a x + 10 a x \equiv 0(\bmod 11)$, meaning $\{a, b\}$ and $\{x, y\}$ cannot both belong to $T$.

Thus, at least one of $T \cap M_{1}$ and $T \cap M_{3}$ is an empty set.

Similarly, at least one of $T \cap M_{2}$ and $T \cap M_{4}$ is an empty set.

Therefore, $T$ can have at most $10 + 10 + 5 = 25$ elements.

On the other hand, take $T = M_{1} \cup M_{2} \cup M_{5}$. Then $T$ has 25 elements.

For any two different elements $\{a, b\}, \{x, y\}$ in $T$, assume
$$
k a \equiv b(\bmod 11), l x \equiv y(\bmod 11),
$$
where $k, l \in \{2,3,10\}$.
Then $a x + b y \equiv (k l + 1) a x(\bmod 11)$,
$a y + b x \equiv (k + l) a x(\bmod 11)$.
It is easy to verify that $k l + 1$ and $k + l$ cannot be divisible by 11.
Thus, $11 \nmid (a x + b y)(a y + b x)$, meaning $T$ satisfies the condition.

In conclusion, the maximum number of elements in $T$ is 25.