Example 13 (1993 Shanghai Competition Question) Given real numbers $x_{1}, x_{2}, x_{3}$ satisfy the equations $x_{1}+\frac{1}{2} x_{2}+\frac{1}{3} x_{3}=$ 1 and $x_{1}^{2}+\frac{1}{2} x_{2}^{2}+\frac{1}{3} x_{3}^{2}=3$, then what is the minimum value of $x_{3}$?
Solve the equations which can be transformed into $x_{1}+\frac{1}{2} x_{2}=1-\frac{1}{3} x_{3}, x_{1}^{2}+\frac{1}{2} x_{2}^{2}=3-\frac{1}{3} x_{3}^{2}$, and ingeniously set the vectors $\vec{a}=\left(x_{1}, \frac{1}{\sqrt{2}} x_{2}\right), \vec{b}=\left(1, \frac{1}{\sqrt{2}}\right)$, then
$$
3-\frac{1}{3} x_{3}^{2}=x_{1}^{2}+\frac{1}{2} x_{2}^{2}=|\vec{a}|^{2} \geqslant \frac{(\vec{a} \cdot \vec{b})^{2}}{|\vec{b}|^{2}}=\frac{\left(x_{1}+\frac{1}{2} x_{2}\right)^{2}}{1+\frac{1}{2}}=\frac{2}{3}\left(1-\frac{1}{3} x_{3}\right)^{2} \text {, }
$$

Solve $3-\frac{1}{3} x_{3}^{2} \geqslant \frac{2}{3}\left(1-\frac{1}{3} x_{3}\right)^{2}$, we get $-\frac{21}{11} \leqslant x_{3} \leqslant 3$, then the minimum value of $x_{3}$ is $-\frac{21}{11}$.