3. In a regular quadrilateral pyramid $P-ABCD$, it is known that $PA=5, AB$ $=6, M$ is the centroid of $\triangle PAD$. Then the volume of the tetrahedron $MPBC$ is $\qquad$ .
3. $4 \sqrt{7}$.

As shown in Figure 2, let $PM$ intersect $AD$ at point $N$. Then $PN=\sqrt{PA^{2}-\left(\frac{AD}{2}\right)^{2}}=4$. Let $PH$ be the height of the pyramid $P-ABCD$. Then $PH=\sqrt{PN^{2}-\left(\frac{AB}{2}\right)^{2}}=\sqrt{7}$. Since $S_{\triangle H}$ shape $ABCD=36$, therefore, $S_{\triangle \triangle BCC}=18$. Hence, $V_{\text {tetrahedron } PABC}=\frac{1}{3} \times 18 \times \sqrt{7}=6 \sqrt{7}$. Also, $\frac{PM}{PN}=\frac{2}{3}$, thus,