$\text{ The equation }16x^2-3y^2-32x+12y-44=0\text{ represents a hyperbola}$

$\text{ Let the equation of any normal be }y=-tx+2t+t^3\text{. }$

$\text{ Since it passes through the point }(15,12)\text{, we have }$

$\begin{array}{r}12=-15t+2t+t^3\\ \text{ or }{t}^3-13t-12=0\end{array}$

$\text{ One root is }-1\text{. Then, }$

$\begin{array}{r}(t+1)\left(t^2+t-12\right)=0\\ \text{ or }t=1,3,4\end{array}$

$\text{ Therefore, the co-normal points are }(1,-2),(9,-6)\text{, and }(16,8)\text{. }$

$\text{ Therefore, the centroid is }(26/3,0)\text{. }$