Example 2 If $w=\sqrt{2 p-q}+\sqrt{3 q-2 p}+\sqrt{6-2 q}$, where $p, q$ are real numbers that make $w$ meaningful, try to determine the maximum value of $w$.

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Consider the function $y=\sqrt{x}$ is convex on $[0,+\infty)$. And $2 p-q \geqslant 0,3 q-2 p \geqslant 0$, $6-2 q \geqslant 0$, and $2 p-q, 3 q-2 p, 6-2 q$ are within $[0,+\infty)$, with their sum being a constant 6. Therefore, when $2 p-q=3 q-2 p=6-2 q=\frac{6}{3}=2$,
that is, when $p=q=2$, $w$ reaches its maximum value, which is
$$
w_{\text {max }}=3 \sqrt{\frac{1}{3}[(2 p-q)+(3 q-2 p)+(6-2 q)}=3 \sqrt{2} \text {. }
$$