1. The problem asks us to find the equation of a line parallel to the line given by $$y = -\frac{1}{3}x + 1$$ that passes through the point $(1, -2)$.\n\n2. Important rule: Parallel lines have the same slope. So, the slope of the new line will be the same as the original line, which is $$m = -\frac{1}{3}$$.\n\n3. Use the point-slope form of a line equation: $$y - y_1 = m(x - x_1)$$ where $(x_1, y_1)$ is the point the line passes through and $m$ is the slope.\n\n4. Substitute $m = -\frac{1}{3}$ and the point $(1, -2)$ into the formula:\n$$y - (-2) = -\frac{1}{3}(x - 1)$$\nwhich simplifies to\n$$y + 2 = -\frac{1}{3}x + \frac{1}{3}$$\n\n5. Solve for $y$ to get the equation in slope-intercept form:\n$$y = -\frac{1}{3}x + \frac{1}{3} - 2$$\n$$y = -\frac{1}{3}x - \frac{5}{3}$$\n\n6. Therefore, the equation of the line parallel to $$y = -\frac{1}{3}x + 1$$ and passing through $(1, -2)$ is $$\boxed{y = -\frac{1}{3}x - \frac{5}{3}}$$.