11. B. Let real numbers $a, b$ satisfy
$$
3 a^{2}-10 a b+8 b^{2}+5 a-10 b=0 \text {. }
$$

Find the minimum value of $u=9 a^{2}+72 b+2$.
11. B. From
$$
3 a^{2}-10 a b+8 b^{2}+5 a-10 b=0 \text {, }
$$

we get $(a-2 b)(3 a-4 b+5)=0$.
Therefore, $a-2 b=0$ or $3 a-4 b+5=0$.
(1) When $a-2 b=0$,
$$
\begin{array}{l}
u=9 a^{2}+72 b+2=36 b^{2}+72 b+2 \\
=36(b+1)^{2}-34 .
\end{array}
$$

Thus, when $b=-1$, the minimum value of $u$ is -34.
(2) When $3 a-4 b+5=0$,
$$
\begin{array}{l}
\quad u=9 a^{2}+72 b+2=16 b^{2}+32 b+27 \\
=16(b+1)^{2}+11 .
\end{array}
$$

Thus, when $b=-1$, the minimum value of $u$ is 11.
In summary, the minimum value of $u$ is -34.