Let $$x$$ and $$y$$ be real numbers such that $$50\left(\frac{2x}{1+3i} - \frac{y}{1-2i}\right) = 31 + 17i$$, $$i = \sqrt{-1}$$. Then the value of $$10(x - 3y)$$ is :

We rationalize each fraction. For the first term:

$$\frac{2x}{1+3i} = \frac{2x(1-3i)}{(1+3i)(1-3i)} = \frac{2x(1-3i)}{1+9} = \frac{x(1-3i)}{5}$$

For the second term:

$$\frac{y}{1-2i} = \frac{y(1+2i)}{(1-2i)(1+2i)} = \frac{y(1+2i)}{1+4} = \frac{y(1+2i)}{5}$$

Substituting into the given equation:

$$50\left(\frac{x(1-3i)}{5} - \frac{y(1+2i)}{5}\right) = 31 + 17i$$

$$10\left(x(1-3i) - y(1+2i)\right) = 31 + 17i$$

$$10\left((x - y) + i(-3x - 2y)\right) = 31 + 17i$$

Comparing real and imaginary parts:

$$10(x - y) = 31 \quad \Rightarrow \quad x - y = \frac{31}{10} \quad \ldots (1)$$

$$10(-3x - 2y) = 17 \quad \Rightarrow \quad -30x - 20y = 17 \quad \ldots (2)$$

From equation (1), $$x = y + \frac{31}{10}$$. Substituting into equation (2):

$$-30\left(y + \frac{31}{10}\right) - 20y = 17$$

$$-30y - 93 - 20y = 17$$

$$-50y = 110$$

$$y = -\frac{11}{5}$$

So $$x = -\frac{11}{5} + \frac{31}{10} = \frac{-22 + 31}{10} = \frac{9}{10}$$.

Now we compute:

$$10(x - 3y) = 10\left(\frac{9}{10} - 3\left(-\frac{11}{5}\right)\right) = 10\left(\frac{9}{10} + \frac{33}{5}\right) = 10\left(\frac{9 + 66}{10}\right) = 75$$

Hence, the correct answer is Option 4.