Solution:

We use Stewart's Theorem with the median $AD$ (note that $BD = CD = \frac{BC}{2}$):
$$CA^{2}(BD) + AB^{2}(CD) = BC\left(AD^{2} + (BD)(CD)\right) \rightarrow \frac{1}{2}\left(CA^{2} + AB^{2}\right) = AD^{2} + \left(\frac{BC}{2}\right)^{2}$$
Similarly, by using Stewart's Theorem with the medians $BE$ and $CF$, we get:
$$\frac{1}{2}\left(BC^{2} + AB^{2}\right) = BE^{2} + \left(\frac{CA}{2}\right)^{2}$$
and
$$\frac{1}{2}\left(CA^{2} + BC^{2}\right) = CF^{2} + \left(\frac{AB}{2}\right)^{2}$$
Adding these three equations, we have:
$$AB^{2} + BC^{2} + CA^{2} = AD^{2} + BE^{2} + CF^{2} + \frac{1}{4}\left(AB^{2} + BC^{2} + CA^{2}\right)$$
Thus,
$$AD^{2} + BE^{2} + CF^{2} = \frac{3}{4}\left(AB^{2} + BC^{2} + CA^{2}\right) = \frac{3}{4}\left(6^{2} + 10^{2} + 14^{2}\right) = 249$$