4. Given the tetrahedron $ABCD$ with the four faces $\triangle DBC$, $\triangle DCA$, $\triangle DAB$, and $\triangle ABC$ having areas of 12, 21, 28, and 37 respectively, and the distance from vertex $D$ to the plane $\triangle ABC$ is $h$. Then $h=$ $\qquad$
4. $\frac{504 \sqrt{2}}{37}$.

Notice that $12^{2}+21^{2}+28^{2}=37^{2}$.
Therefore, the tetrahedron $A B C D$ is a right-angled tetrahedron.
$$
\begin{array}{l}
\text { Hence } h=\frac{3 D A \cdot D B \cdot D C}{S_{\triangle A B C}} \\
=\frac{3 \sqrt{24 \times 42 \times 56}}{37}=\frac{504 \sqrt{2}}{37} .
\end{array}
$$