1. The problem asks for the equation of a line perpendicular to $y=\frac{1}{3}x - 5$ that passes through the point $(-2, 12)$.\n\n2. Recall the slope-intercept form is $y=mx+b$, where $m$ is the slope. The given line has slope $m=\frac{1}{3}$.\n\n3. The slope of a line perpendicular to another is the negative reciprocal of the original slope. So, the perpendicular slope $m_\perp = -\frac{1}{\frac{1}{3}} = -3$.\n\n4. The point-slope form of a line is given by $$y - y_1 = m(x - x_1)$$ where $(x_1, y_1)$ is a point on the line and $m$ is the slope.\n\n5. Substitute $m=-3$ and the point $(-2, 12)$ into the point-slope form: $$y - 12 = -3(x - (-2))$$\n\n6. Simplify the expression inside the parentheses: $$y - 12 = -3(x + 2)$$\n\n7. This is the equation of the line in point-slope form that is perpendicular to the given line and passes through $(-2, 12)$.