Example 14 Given real numbers $a, b, c, d, e$ satisfy $3a + 2b - c + 4d + \sqrt{133}e = \sqrt{133}, 2a^2 + 3b^2 + 3c^2 + d^2 + 6e^2 = 60$, try to find when $e$ takes its extreme values, and what are the maximum and minimum values respectively.
The given conditions can be rewritten as: $\left\{\begin{array}{l}3 a+2 b-c+4 d=\sqrt{133}-\sqrt{133} e, \\ 2 a^{2}+3 b^{2}+3 c^{2}+d^{2}=60-6 e^{2},\end{array}\right.$
Square both sides of (1) and apply the Cauchy-Schwarz inequality along with (2) to get
Square both sides of (1) and apply the Cauchy-Schwarz inequality along with (2) to get
$$
\begin{array}{c}
(\sqrt{133}-\sqrt{133} e)^{2}=(3 a+2 b-c+4 d)^{2}=\left(\frac{3}{\sqrt{2}} \cdot \sqrt{2} a+\frac{2}{\sqrt{3}} \cdot \sqrt{3} b+\frac{-1}{\sqrt{3}} \cdot \sqrt{3} c+4 d\right)^{2} \leqslant \\
\left(\frac{9}{2}+\frac{4}{3}+\frac{1}{3}+16\right) \cdot\left(2 a^{2}+3 b^{2}+3 c^{2}+d^{2}\right)=\frac{133}{6}\left(60-6 e^{2}\right)=133\left(10-e^{2}\right) .
\end{array}
$$

Simplifying, we get $2 e^{2}-2 e-9 \leqslant 0$, solving which yields $\quad \frac{1-\sqrt{19}}{2} \leqslant e \leqslant \frac{1+\sqrt{19}}{2}$.
The equality in the inequality $2 e^{2}-2 e-9 \leqslant 0$ holds if and only if $\frac{2 a}{3}=\frac{3 b}{2}=-3 c=\frac{d}{4}$, which means $e$ can only achieve its maximum and minimum values under this condition.

From $\frac{2 a}{3}=\frac{3 b}{2}=-3 c=\frac{d}{4}$, we get $a=\frac{3 d}{8}, b=\frac{d}{6}, c=-\frac{d}{12}$, substituting these into $3 a+2 b-c+4 d=\sqrt{133}-\sqrt{133} e$ and $2 a^{2}+3 b^{2}+3 c^{2}+d^{2}=60-6 e^{2}$, we obtain
$$
d=\frac{24(1-e)}{\sqrt{133}}, d^{2}=\frac{96\left(60-6 e^{2}\right)}{133} .
$$

Solving the system of equations $\left\{\begin{array}{l}d=\frac{24(1-e)}{\sqrt{133}}, \\ d^{2}=\frac{96\left(60-6 e^{2}\right)}{133},\end{array}\right.$
we get $\left\{\begin{array}{l}e_{1}=\frac{1-\sqrt{19}}{2} \\ d_{1}=\frac{12(\sqrt{133}+19 \sqrt{7})}{133}\end{array},\left\{\begin{array}{l}e_{2}=\frac{1+\sqrt{19}}{2} \\ d_{2}=\frac{12(\sqrt{133}-19 \sqrt{7})}{133}\end{array}\right.\right.$.
Therefore, when $a=\frac{3 d_{1}}{8}, b=\frac{d_{1}}{6}, c=-\frac{d_{1}}{12}, d=d_{1}$, $e$ takes its minimum value $\frac{1-\sqrt{19}}{2}$,
and when $a=\frac{3 d_{2}}{8}, b=\frac{d_{2}}{6}, c=-\frac{d_{2}}{12}, d=d_{2}$, $e$ takes its maximum value $\frac{1+\sqrt{19}}{2}$.