Let $$ABCD$$ be a triangle formed by the lines $$7x − 6y + 3 = 0, x + 2y − 31 = 0$$ and $$9x − 2y − 19 = 0.$$ Let the point $$(h,k)$$ be the image of the centroid of $$\triangle ABC$$ in the line $$3x + 6y − 53 = 0.$$ Then $$h^{2}+k^{2}+hk$$ is equal to :

The vertices of triangle ABC are found by solving the pairs of equations of the lines.

Intersection of $$x + 2y - 31 = 0$$ and $$9x - 2y - 19 = 0$$:
Adding the equations: $$x + 2y - 31 + 9x - 2y - 19 = 0$$ → $$10x - 50 = 0$$ → $$x = 5$$.
Substituting $$x = 5$$ into $$x + 2y - 31 = 0$$: $$5 + 2y - 31 = 0$$ → $$2y = 26$$ → $$y = 13$$.
Thus, vertex A is $$(5, 13)$$.

Intersection of $$7x - 6y + 3 = 0$$ and $$9x - 2y - 19 = 0$$:
Multiplying the first equation by 1 and the second by 3: $$7x - 6y = -3$$ and $$27x - 6y = 57$$.
Subtracting the first from the second: $$(27x - 6y) - (7x - 6y) = 57 - (-3)$$ → $$20x = 60$$ → $$x = 3$$.
Substituting $$x = 3$$ into $$7x - 6y + 3 = 0$$: $$21 - 6y + 3 = 0$$ → $$24 = 6y$$ → $$y = 4$$.
Thus, vertex B is $$(3, 4)$$.

Intersection of $$7x - 6y + 3 = 0$$ and $$x + 2y - 31 = 0$$:
Multiplying the second equation by 3: $$3x + 6y - 93 = 0$$.
Adding to the first equation: $$(7x - 6y + 3) + (3x + 6y - 93) = 0$$ → $$10x - 90 = 0$$ → $$x = 9$$.
Substituting $$x = 9$$ into $$x + 2y - 31 = 0$$: $$9 + 2y - 31 = 0$$ → $$2y = 22$$ → $$y = 11$$.
Thus, vertex C is $$(9, 11)$$.

The centroid G of triangle ABC is the average of the vertices' coordinates:
x-coordinate: $$\frac{5 + 3 + 9}{3} = \frac{17}{3}$$
y-coordinate: $$\frac{13 + 4 + 11}{3} = \frac{28}{3}$$
Thus, G is $$\left(\frac{17}{3}, \frac{28}{3}\right)$$.

The image of G in the line $$3x + 6y - 53 = 0$$ is $$(h, k)$$. The formula for the image of a point $$(x_1, y_1)$$ in the line $$ax + by + c = 0$$ is given by:
$$\frac{h - x_1}{a} = \frac{k - y_1}{b} = \frac{-2(ax_1 + by_1 + c)}{a^2 + b^2}$$
Here, $$a = 3$$, $$b = 6$$, $$c = -53$$, $$x_1 = \frac{17}{3}$$, $$y_1 = \frac{28}{3}$$.

First, compute $$ax_1 + by_1 + c$$:
$$3 \cdot \frac{17}{3} + 6 \cdot \frac{28}{3} - 53 = 17 + 56 - 53 = 20$$

Then, $$a^2 + b^2 = 3^2 + 6^2 = 9 + 36 = 45$$.

So,
$$\frac{h - \frac{17}{3}}{3} = \frac{k - \frac{28}{3}}{6} = \frac{-2 \cdot 20}{45} = \frac{-40}{45} = -\frac{8}{9}$$

Solving for $$h$$:
$$\frac{h - \frac{17}{3}}{3} = -\frac{8}{9}$$
Multiply both sides by 3:
$$h - \frac{17}{3} = -\frac{8}{9} \cdot 3 = -\frac{24}{9} = -\frac{8}{3}$$
$$h = -\frac{8}{3} + \frac{17}{3} = \frac{9}{3} = 3$$

Solving for $$k$$:
$$\frac{k - \frac{28}{3}}{6} = -\frac{8}{9}$$
Multiply both sides by 6:
$$k - \frac{28}{3} = -\frac{8}{9} \cdot 6 = -\frac{48}{9} = -\frac{16}{3}$$
$$k = -\frac{16}{3} + \frac{28}{3} = \frac{12}{3} = 4$$

Thus, $$(h, k) = (3, 4)$$.

Now compute $$h^2 + k^2 + hk$$:
$$3^2 + 4^2 + 3 \cdot 4 = 9 + 16 + 12 = 37$$

The value is 37, which corresponds to option B.