jmc

If $D$ is the centroid of triangle $ABC$, then $ABD$, $ACD$, and $BCD$ would all have equal areas (to see this, remember that the medians of a triangle divide the triangle into 6 equal areas). There is only one point with this property (if we move around $D$, the area of one of the small triangles will increase and will no longer be $1/3$ of the total area). So $D$ must be the centroid of triangle $ABC$. The $x$ and $y$ coordinates of the centroid are found by averaging the $x$ and $y$ coordinates, respectively, of the vertices, so $(m,n)=\left(\frac{5+3+6}{3},\frac{8+(-2)+1}{3}\right)=\left(\frac{14}{3},\frac{7}{3}\right)$, and $10m+n=10\left(\frac{14}{3}\right)+\frac{7}{3}=\boxed{49}$.