The foot of the perpendicular from the origin to a plane is $(12,-4,3).$  Find the equation of the plane.  Enter your answer in the form
\[Ax + By + Cz + D = 0,\]where $A,$ $B,$ $C,$ $D$ are integers such that $A > 0$ and $\gcd(|A|,|B|,|C|,|D|) = 1.$
We can take $\begin{pmatrix} 12 \\ -4 \\ 3 \end{pmatrix}$ as the normal vector of the plane.  Then the equation of the plane is of the form
\[12x - 4y + 3z + D = 0.\]Substituting in the coordinates of $(12,-4,3),$ we find that the equation of the plane is $\boxed{12x - 4y + 3z - 169 = 0}.$
We can take $\begin{pmatrix} 12 \\ -4 \\ 3 \end{pmatrix}$ as the normal vector of the plane.  Then the equation of the plane is of the form
\[12x - 4y + 3z + D = 0.\]Substituting in the coordinates of $(12,-4,3),$ we find that the equation of the plane is $\boxed{12x - 4y + 3z - 169 = 0}.$
The final answer is $The final answer is $\boxed{12x - 4y + 3z - 169$. I hope it is correct.