3. Let $x+y+z=1$. Then the minimum value of the function $u=2 x^{2}+3 y^{2}$ $+z^{2}$ is . $\qquad$
3. $\frac{6}{11}$.

From the given conditions and Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
1=x+y+z=\left(\frac{1}{\sqrt{2}} \cdot \sqrt{2} x+\frac{1}{\sqrt{3}} \cdot \sqrt{3} y+1 \cdot z\right) \\
\leqslant\left(\frac{1}{2}+\frac{1}{3}+1\right)^{\frac{1}{2}}\left(2 x^{2}+3 y^{2}+z^{2}\right)^{\frac{1}{2}} \\
=\sqrt{\frac{11}{6}} \times \sqrt{u} .
\end{array}
$$

Thus, $u \geqslant \frac{6}{11}$.
The equality in (1) holds if and only if
$$
\frac{\sqrt{2} x}{\frac{1}{\sqrt{2}}}=\frac{\sqrt{3} y}{\frac{1}{\sqrt{3}}}=\frac{z}{1}=\lambda \text {, }
$$

which implies $x=\frac{\lambda}{2}, y=\frac{\lambda}{3}, z=\lambda$.
Substituting into the given equation, we get $\frac{\lambda}{2}+\frac{\lambda}{3}+\lambda=1$, solving for $\lambda$ yields $\lambda=\frac{6}{11}$.
Therefore, when $x=\frac{3}{11}, y=\frac{2}{11}, z=\frac{6}{11}$, $u_{\text {min }}=\frac{6}{11}$.