1. **State the problem:** Find the farthest distance from the point $(12, 2)$ to the circle given by the equation $$x^2 + y^2 + 6x - 16y + 24 = 0.$$\n\n2. **Rewrite the circle equation in standard form:** Complete the square for $x$ and $y$.\n\nFor $x$: $$x^2 + 6x = (x+3)^2 - 9.$$\nFor $y$: $$y^2 - 16y = (y-8)^2 - 64.$$\n\nSubstitute back: $$ (x+3)^2 - 9 + (y-8)^2 - 64 + 24 = 0.$$\nSimplify constants: $$ (x+3)^2 + (y-8)^2 - 49 = 0.$$\nSo, $$ (x+3)^2 + (y-8)^2 = 49.$$\n\n3. **Identify the center and radius:**\nCenter $C = (-3, 8)$\nRadius $r = \sqrt{49} = 7.$\n\n4. **Calculate the distance from the point $P = (12, 2)$ to the center $C$:**\n$$ d = \sqrt{(12 + 3)^2 + (2 - 8)^2} = \sqrt{15^2 + (-6)^2} = \sqrt{225 + 36} = \sqrt{261}.$$\n\n5. **Find the farthest distance from $P$ to the circle:**\nThe farthest distance is the distance from $P$ to the center plus the radius: $$ d_{max} = d + r = \sqrt{261} + 7.$$\n\n6. **Final answer:**\n$$\boxed{\sqrt{261} + 7}.$$