3. Given that $x$, $y$, and $z$ are real numbers, and
$$
x^{2}+4 y^{2}+9 z^{2}=2009 \text{. }
$$

Then the maximum value of $x+2 y+3 z$ is $\qquad$
$3.7 \sqrt{123}$.
By the AM-GM inequality, we have
$$
\frac{x+2 y+3 z}{3} \leqslant \sqrt{\frac{x^{2}+4 y^{2}+9 z^{2}}{3}}=\sqrt{\frac{2009}{3}} \text {. }
$$

Thus, $x+2 y+3 z \leqslant 7 \sqrt{123}$.
Equality holds if and only if $x=2 y=3 z=\sqrt{\frac{2009}{3}}$.