Example 11. Let $\mathrm{a}$ be a real number, find the minimum value of the quadratic function
$$
y=x^{2}-4 a x+5 a^{2}-3 a
$$

denoted as $\mathrm{m}$.
When $a$ varies in $0 \leqslant a^{2}-4 a-2 \leqslant 10$, find the maximum value of $m$.
$$
\text{Solve} \begin{aligned}
y & = x^{2} - 4ax + 5a^{2} - 3a \\
& = (x - 2a)^{2} + a^{2} - 3a
\end{aligned}
$$

The above expression achieves its minimum value when $x = 2a$, so
$$
m = a^{2} - 3a = \left(a - \frac{3}{2}\right)^{2} - \frac{9}{4}.
$$

Furthermore, $0 \leqslant a^{2} - 4a - 2 \leqslant 10$ implies
$$
0 \leqslant a^{2} - 4a - 2,
$$

and $10 \geqslant x^{2} - 4a - 12$.
From (1), we get $a \leqslant 2 - \sqrt{6}$, or $a \geqslant 2 + \sqrt{6}$.
From (2), we get $-2 \leqslant a \leqslant 6$.
From (3) and (4), the range of $a$ that satisfies both (1) and (2) is $-2 \leqslant a \leqslant 2 - \sqrt{6}$ or $2 + \sqrt{6} \leqslant a \leqslant 6$.
In the above range, when $a = 6$, $\left|a - \frac{3}{2}\right|$ is maximized.
Therefore, when $a = 6$, $m$ reaches its maximum value of 18.
$$