jmc

First, we know that this point is above the $x$-axis because it is closer to a point in the first quadrant than it is to the $x$-axis. Next, we know that $y=12$ from the given information. By the distance formula, we have the equation $\sqrt{(x-1{)}^2+(12-6{)}^2}=10$. Solving, we have $$\begin{array}{rl}\sqrt{(x-1{)}^2+(12-6{)}^2}=10& \\ x^2-2x+1+36& =100\\ x^2-2x-63& =0\\ (x-9)(x+7)& =0\end{array}$$Thus, $x-9=0$ or $x+7=0$, so $x=9$ or $x=-7$. $x=9$ by the given conditions. Thus, our point is $(9,12)$ and is a distance of $\sqrt{9^2+12^2}=15$ units from the origin. $n=\boxed{15}$.