let's first off the slope of 3y = x - 6, by solving for "y" and therefore putting it in slope-intercept form.
[tex]\bf 3y=x-6\implies y=\cfrac{x-6}{3}\implies y=\cfrac{1}{3}x-\cfrac{6}{3} \\\\\\ y=\cfrac{1}{3}x-2\qquad \impliedby \qquad \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array}[/tex]
[tex]\bf ~\dotfill\\\\ \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{1}{3}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{3}{1}}\qquad \stackrel{negative~reciprocal}{-\cfrac{3}{1}\implies -3}}[/tex]
so we're really looking for the equation of a line whose slope is -3 and runs through 7, -7.
[tex]\bf (\stackrel{x_1}{7}~,~\stackrel{y_1}{-7})~\hspace{10em} slope = m\implies -3 \\\\\\ \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-(-7)=-3(x-7) \\\\\\ y+7=-3x+21\implies \stackrel{y}{f(x)}=-3x+14[/tex]
