10. Let positive real numbers $a, b, c$ satisfy $a^{2}+4 b^{2}+9 c^{2}=4 b+12 c-2$, find the minimum value of $\frac{1}{a}+\frac{2}{b}+\frac{3}{c}$.
$$
a^{2}+4 b^{2}+9 c^{2}=4 b+12 c-2 \Rightarrow a^{2}+(2 b-1)^{2}+(3 c-2)^{2}=3,
$$

By the Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
(1+1+1)\left[a^{2}+(2 b-1)^{2}+(3 c-2)^{2}\right] \geqslant(a+2 b+3 c-3)^{2} \\
\Rightarrow(a+2 b+3 c-3)^{2} \leqslant 9 \Rightarrow 0<a+2 b+3 c \leqslant 6 .
\end{array}
$$

Then, by the weighted AM-HM inequality, we get
$$
\frac{1}{a}+\frac{2}{b}+\frac{3}{c}=\frac{1}{a}+\frac{4}{2 b}+\frac{9}{3 c} \geqslant \frac{(1+2+3)^{2}}{a+2 b+3 c}=\frac{36}{a+2 b+3 c} \geqslant \frac{36}{6}=6,
$$

Equality holds when $a=b=c=1$.
Therefore, the minimum value of $\frac{1}{a}+\frac{2}{b}+\frac{3}{c}$ is 6.