11. (20 points) Natural numbers $a, b$ make $4a + 7b$ and $5a + 6b$ both multiples of 11, and $a + 21b \geq 792$. Find the minimum value of $T = 21a + b$.
11. Let $4a + 7b = 11x$, $5a + 6b = 11y$. Then
$$
\begin{array}{l}
a = 7y - 6x \geqslant 0, \\
b = 5x - 4y \geqslant 0.
\end{array}
$$

Thus, $T = 11(13y - 11x)$.
Let $13y - 11x = r$.
From $a + 21b \geqslant 792$, we get
$$
9x - 7y \geqslant 72.
$$

Substituting equation (3) into equation (1) to eliminate $y$ gives $x \leqslant 7r$.
(1) + (4) gives $x \geqslant 24$.

Thus, $7r \geqslant 24 \Rightarrow r \geqslant 4$.
Therefore, $T \geqslant 44$.

On the other hand, when $x = 28$, $y = 24$, we have $a = 0$, $b = 44$, $a + 21b = 924 > 792$, and $T = 44$. In summary, $T_{\min} = 44$.