Let AB be a chord of length 12 of the circle $$(x-2)^2 + (y+1)^2 = \frac{169}{4}$$. If tangents drawn to the circle at points A and B intersect at the point P, then five times the distance of point P from chord AB is equal to _____

We have the circle $$(x-2)^2 + (y+1)^2 = \frac{169}{4}$$ with center $$C = (2, -1)$$ and radius $$r = \frac{13}{2}$$. A chord $$AB$$ has length 12, and tangents drawn to the circle at points $$A$$ and $$B$$ intersect at point $$P$$. We need to find five times the distance from $$P$$ to chord $$AB$$.

Let $$M$$ be the midpoint of chord $$AB$$. The perpendicular from the center of a circle to a chord bisects the chord, so $$CM \perp AB$$ and $$AM = MB = 6$$. Using the Pythagorean theorem in right triangle $$CMA$$: $$CM = \sqrt{CA^2 - AM^2} = \sqrt{r^2 - 36} = \sqrt{\frac{169}{4} - 36} = \sqrt{\frac{169 - 144}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$$.

By the symmetry of the configuration ($$A$$ and $$B$$ are symmetric about line $$CM$$), the intersection point $$P$$ of the tangents at $$A$$ and $$B$$ lies on line $$CM$$, on the opposite side of the chord from the center. The distance from $$P$$ to the chord is $$PM$$, and $$CP = CM + PM = \frac{5}{2} + PM$$.

Since $$PA$$ is tangent to the circle at $$A$$, the radius $$CA$$ is perpendicular to the tangent $$PA$$, so $$\angle CAP = 90°$$. In right triangle $$CAP$$, the Pythagorean theorem gives $$CP^2 = CA^2 + PA^2 = \frac{169}{4} + PA^2$$.

Also, since $$PM \perp AB$$ and $$A$$ lies on chord $$AB$$, triangle $$PMA$$ is right-angled at $$M$$. So $$PA^2 = PM^2 + MA^2 = PM^2 + 36$$.

Substituting into the equation from triangle $$CAP$$: $$\left(\frac{5}{2} + PM\right)^2 = \frac{169}{4} + PM^2 + 36$$.

Expanding the left side: $$\frac{25}{4} + 5 \cdot PM + PM^2 = \frac{169}{4} + PM^2 + 36$$.

The $$PM^2$$ terms cancel, leaving $$\frac{25}{4} + 5 \cdot PM = \frac{169}{4} + 36$$. Computing the right side: $$\frac{169}{4} + \frac{144}{4} = \frac{313}{4}$$.

So $$5 \cdot PM = \frac{313}{4} - \frac{25}{4} = \frac{288}{4} = 72$$, giving $$PM = \frac{72}{5}$$.

Therefore, five times the distance from $$P$$ to chord $$AB$$ is $$5 \times PM = 5 \times \frac{72}{5} = 72$$.

Hence, the correct answer is $$\boxed{72}$$.