bearing in mind that perpendicular lines have negative reciprocal slopes, let's find firstly the slope of AC.
[tex]\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{1})\qquad (\stackrel{x_2}{1}~,~\stackrel{y_2}{6}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{6-1}{1-2}\implies \cfrac{5}{-1}\implies -5 \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{-5}{1}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{1}{-5}}\qquad \stackrel{negative~reciprocal}{\cfrac{1}{5}}}[/tex]
so, we're really looking for the equation of a line whose slope is 1/5 and that passes through (3,3)
[tex]\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{3}) ~\hspace{10em}slope = m\implies \cfrac{1}{5} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-3=\cfrac{1}{5}(x-3) \implies y-3=\cfrac{1}{5}x-\cfrac{3}{5} \\\\\\ y=\cfrac{1}{5}x-\cfrac{3}{5}+3\implies y=\cfrac{1}{5}x+\cfrac{12}{5}[/tex]
