4. Given real numbers $x, y, z$ satisfy $x^{2}+2 y^{2}+3 z^{2}=24$.
Then the minimum value of $x+2 y+3 z$ is $\qquad$ .
4. -12 .

By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
(x+2 y+3 z)^{2} \\
=(1 \times x+\sqrt{2} \times \sqrt{2} y+\sqrt{3} \times \sqrt{3} z)^{2} \\
\leqslant\left[1^{2}+(\sqrt{2})^{2}+(\sqrt{3})^{2}\right]\left(x^{2}+2 y^{2}+3 z^{2}\right) \\
\quad=144 .
\end{array}
$$

Therefore, $x+2 y+3 z \geqslant-12$, with equality holding if and only if $x=y=z=-2$.
Hence, the minimum value of $x+2 y+3 z$ is -12.