Let the triangle PQR be the image of the triangle with vertices (1, 3), (3, 1) and (2, 4) in the line $$x+2y=2$$. If the centroid of $$ \triangle PQR $$ is the point $$ (\alpha, \beta) $$, then $$ 15(\alpha - \beta) $$ is equal to :

To find the image of the triangle with vertices (1, 3), (3, 1), and (2, 4) in the line $$x + 2y = 2$$, we first rewrite the line equation as $$x + 2y - 2 = 0$$. Here, $$a = 1$$, $$b = 2$$, and $$c = -2$$.

The formula for the image $$(x_2, y_2)$$ of a point $$(x_1, y_1)$$ in the line $$ax + by + c = 0$$ is given by:

$$ \frac{x_2 - x_1}{a} = \frac{y_2 - y_1}{b} = \frac{-2(a x_1 + b y_1 + c)}{a^2 + b^2} $$

Substituting $$a = 1$$, $$b = 2$$, and $$c = -2$$:

$$ \frac{x_2 - x_1}{1} = \frac{y_2 - y_1}{2} = \frac{-2(1 \cdot x_1 + 2 \cdot y_1 - 2)}{1^2 + 2^2} = \frac{-2(x_1 + 2y_1 - 2)}{5} $$

Let the common ratio be $$k$$, so:

$$ x_2 = x_1 + k, \quad y_2 = y_1 + 2k, \quad \text{where} \quad k = \frac{-2(x_1 + 2y_1 - 2)}{5} $$

Solving for $$x_2$$ and $$y_2$$:

$$ x_2 = x_1 + \frac{-2(x_1 + 2y_1 - 2)}{5} = \frac{5x_1 - 2x_1 - 4y_1 + 4}{5} = \frac{3x_1 - 4y_1 + 4}{5} $$

$$ y_2 = y_1 + 2 \cdot \frac{-2(x_1 + 2y_1 - 2)}{5} = \frac{5y_1 - 4x_1 - 8y_1 + 8}{5} = \frac{-4x_1 - 3y_1 + 8}{5} $$

Now, find the images of each vertex:

Image of A(1, 3):

$$ x_P = \frac{3(1) - 4(3) + 4}{5} = \frac{3 - 12 + 4}{5} = \frac{-5}{5} = -1 $$

$$ y_P = \frac{-4(1) - 3(3) + 8}{5} = \frac{-4 - 9 + 8}{5} = \frac{-5}{5} = -1 $$

So, P is $$(-1, -1)$$.

Image of B(3, 1):

$$ x_Q = \frac{3(3) - 4(1) + 4}{5} = \frac{9 - 4 + 4}{5} = \frac{9}{5} $$

$$ y_Q = \frac{-4(3) - 3(1) + 8}{5} = \frac{-12 - 3 + 8}{5} = \frac{-7}{5} $$

So, Q is $$\left(\frac{9}{5}, -\frac{7}{5}\right)$$.

Image of C(2, 4):

$$ x_R = \frac{3(2) - 4(4) + 4}{5} = \frac{6 - 16 + 4}{5} = \frac{-6}{5} $$

$$ y_R = \frac{-4(2) - 3(4) + 8}{5} = \frac{-8 - 12 + 8}{5} = \frac{-12}{5} $$

So, R is $$\left(-\frac{6}{5}, -\frac{12}{5}\right)$$.

The vertices of $$\triangle PQR$$ are P$$(-1, -1)$$, Q$$\left(\frac{9}{5}, -\frac{7}{5}\right)$$, and R$$\left(-\frac{6}{5}, -\frac{12}{5}\right)$$.

The centroid $$(\alpha, \beta)$$ of a triangle with vertices $$(x_1, y_1)$$, $$(x_2, y_2)$$, $$(x_3, y_3)$$ is given by:

$$ \alpha = \frac{x_1 + x_2 + x_3}{3}, \quad \beta = \frac{y_1 + y_2 + y_3}{3} $$

Compute $$\alpha$$:

$$ \alpha = \frac{-1 + \frac{9}{5} + \left(-\frac{6}{5}\right)}{3} = \frac{-1 + \frac{9}{5} - \frac{6}{5}}{3} $$

First, simplify the numerator:

$$ -1 + \frac{9}{5} - \frac{6}{5} = -1 + \frac{3}{5} = -\frac{5}{5} + \frac{3}{5} = -\frac{2}{5} $$

Then,

$$ \alpha = \frac{-\frac{2}{5}}{3} = -\frac{2}{5} \times \frac{1}{3} = -\frac{2}{15} $$

Compute $$\beta$$:

$$ \beta = \frac{-1 + \left(-\frac{7}{5}\right) + \left(-\frac{12}{5}\right)}{3} = \frac{-1 - \frac{7}{5} - \frac{12}{5}}{3} $$

Simplify the numerator:

$$ -1 - \frac{7}{5} - \frac{12}{5} = -1 - \frac{19}{5} = -\frac{5}{5} - \frac{19}{5} = -\frac{24}{5} $$

Then,

$$ \beta = \frac{-\frac{24}{5}}{3} = -\frac{24}{5} \times \frac{1}{3} = -\frac{24}{15} = -\frac{8}{5} $$

Now compute $$\alpha - \beta$$:

$$ \alpha - \beta = -\frac{2}{15} - \left(-\frac{8}{5}\right) = -\frac{2}{15} + \frac{8}{5} $$

Convert to a common denominator of 15:

$$ \frac{8}{5} = \frac{8 \times 3}{5 \times 3} = \frac{24}{15} $$

So,

$$ \alpha - \beta = -\frac{2}{15} + \frac{24}{15} = \frac{22}{15} $$

Now compute $$15(\alpha - \beta)$$:

$$ 15(\alpha - \beta) = 15 \times \frac{22}{15} = 22 $$

Therefore, $$15(\alpha - \beta) = 22$$, which corresponds to option D.