2. If $a, b, c > 0, \frac{1}{a}+\frac{2}{b}+\frac{3}{c}=1$, then the minimum value of $a+2b+3c$ is . $\qquad$
2. 36 .

By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
a+2 b+3 c=(a+2 b+3 c)\left(\frac{1}{a}+\frac{2}{b}+\frac{3}{c}\right) \\
\geqslant(1+2+3)^{2}=36 .
\end{array}
$$