The number of common tangents, to the circles $$x^2 + y^2 - 18x - 15y + 131 = 0$$ and $$x^2 + y^2 - 6x - 6y - 7 = 0$$, is

Circle 1: $$x^2 + y^2 - 18x - 15y + 131 = 0$$

Center: $$(9, 7.5)$$, radius: $$r_1 = \sqrt{81 + 56.25 - 131} = \sqrt{6.25} = 2.5$$

Circle 2: $$x^2 + y^2 - 6x - 6y - 7 = 0$$

Center: $$(3, 3)$$, radius: $$r_2 = \sqrt{9 + 9 + 7} = \sqrt{25} = 5$$

Distance between centers:

$$d = \sqrt{(9-3)^2 + (7.5-3)^2} = \sqrt{36 + 20.25} = \sqrt{56.25} = 7.5$$

Now: $$r_1 + r_2 = 2.5 + 5 = 7.5 = d$$

Since $$d = r_1 + r_2$$, the circles are externally tangent.

For externally tangent circles, the number of common tangents = 3 (2 external + 1 internal at the point of tangency).

This matches option 1: 3.