Let the mid points of the sides of a triangle ABC be $$\left(\frac{5}{2}, 7\right)$$, $$\left(\frac{5}{2}, 3\right)$$ and $$(4, 5)$$. If its incentre is $$(h, k)$$, then $$3h + k$$ is equal to :

We are given the midpoints of the sides of triangle $$ABC$$ as $$M_1 = \left(\frac{5}{2}, 7\right)$$, $$M_2 = \left(\frac{5}{2}, 3\right)$$, and $$M_3 = (4, 5)$$. Using the property that each vertex equals the sum of two adjacent midpoints minus the opposite midpoint:

$$A = M_1 + M_3 - M_2 = \left(\frac{5}{2} + 4 - \frac{5}{2},\; 7 + 5 - 3\right) = (4, 9)$$

$$B = M_1 + M_2 - M_3 = \left(\frac{5}{2} + \frac{5}{2} - 4,\; 7 + 3 - 5\right) = (1, 5)$$

$$C = M_2 + M_3 - M_1 = \left(\frac{5}{2} + 4 - \frac{5}{2},\; 3 + 5 - 7\right) = (4, 1)$$

We verify: midpoint of $$AB = \left(\frac{4+1}{2}, \frac{9+5}{2}\right) = \left(\frac{5}{2}, 7\right) = M_1$$, midpoint of $$BC = \left(\frac{1+4}{2}, \frac{5+1}{2}\right) = \left(\frac{5}{2}, 3\right) = M_2$$, midpoint of $$AC = \left(\frac{4+4}{2}, \frac{9+1}{2}\right) = (4, 5) = M_3$$.

Now we find the side lengths:

$$a = BC = \sqrt{(4-1)^2 + (1-5)^2} = \sqrt{9 + 16} = 5$$

$$b = AC = \sqrt{(4-4)^2 + (1-9)^2} = \sqrt{64} = 8$$

$$c = AB = \sqrt{(1-4)^2 + (5-9)^2} = \sqrt{9 + 16} = 5$$

The incenter is given by $$(h, k) = \left(\frac{a \cdot x_A + b \cdot x_B + c \cdot x_C}{a + b + c},\; \frac{a \cdot y_A + b \cdot y_B + c \cdot y_C}{a + b + c}\right)$$:

$$h = \frac{5(4) + 8(1) + 5(4)}{5 + 8 + 5} = \frac{20 + 8 + 20}{18} = \frac{48}{18} = \frac{8}{3}$$

$$k = \frac{5(9) + 8(5) + 5(1)}{18} = \frac{45 + 40 + 5}{18} = \frac{90}{18} = 5$$

Therefore, $$3h + k = 3 \times \frac{8}{3} + 5 = 8 + 5 = 13$$.

Hence, the correct answer is Option 3.