The distance from a vertex of the base of a regular tetrahedron to the centroid of the opposite side is 4. The maximum volume of this regular tetrahedron is
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11. 18.

In a regular tetrahedron \( P-ABC \), let the centroid of the lateral face \( PBC \) be \( G \). Extend \( PG \) to intersect \( BC \) at \( D \). Clearly, \( D \) is the midpoint of \( BC \). Let the projections of \( P \) and \( G \) on the base be \( O \) and \( H \), respectively. The side length of the base is \( a \), and the height of the tetrahedron is \( H \).

Since \( AG = 4 \), and \( GH = \frac{1}{3} PO = \frac{h}{3}, HD = \frac{1}{3} OD = \frac{1}{9} AD = \frac{\sqrt{3}}{18} a \), then \( AH = \frac{4 \sqrt{3}}{9} a \). Therefore, \(\left(\frac{4 \sqrt{3}}{9} a\right)^{2} + \left(\frac{h}{3}\right)^{2} = 4^{2}\), which simplifies to \( 16 a^{2} + 3 h^{2} = 16 \times 27, 8 a^{2} + 8 a^{2} + 3 h^{2} = 16 \times 27. 16 \times 27 \geqslant 3^{3} \sqrt{8 a^{2} \cdot 8 a^{2} \cdot 3 h^{2}}, a^{2} h \leqslant \frac{216}{\sqrt{3}} \), hence \( V_{P-ABC} = \frac{1}{3} \cdot \frac{\sqrt{3}}{4} a^{2} \cdot h \leqslant \frac{\sqrt{3}}{12} \times \frac{216}{\sqrt{3}} = 18 \).