Let $$(\alpha, \beta)$$ be the centroid of the triangle formed by the lines $$15x - y = 82$$, $$6x - 5y = -4$$ and $$9x + 4y = 17$$. Then $$\alpha + 2\beta$$ and $$2\alpha - \beta$$ are the roots of the equation

Find the vertices of the triangle by solving pairs of lines:

Intersection of $$15x - y = 82$$ and $$6x - 5y = -4$$:

From first: $$y = 15x - 82$$. Substituting: $$6x - 5(15x-82) = -4$$

$$6x - 75x + 410 = -4$$

$$-69x = -414$$

$$x = 6, \quad y = 90 - 82 = 8$$. Point: $$(6, 8)$$

Intersection of $$15x - y = 82$$ and $$9x + 4y = 17$$:

$$y = 15x - 82$$. $$9x + 4(15x-82) = 17$$

$$9x + 60x - 328 = 17$$

$$69x = 345$$

$$x = 5, \quad y = 75 - 82 = -7$$. Point: $$(5, -7)$$

Intersection of $$6x - 5y = -4$$ and $$9x + 4y = 17$$:

Multiply first by 4: $$24x - 20y = -16$$

Multiply second by 5: $$45x + 20y = 85$$

Adding: $$69x = 69$$, $$x = 1$$. $$6 - 5y = -4$$, $$y = 2$$. Point: $$(1, 2)$$

Centroid: $$(\alpha, \beta) = \left(\frac{6+5+1}{3}, \frac{8-7+2}{3}\right) = (4, 1)$$

$$\alpha + 2\beta = 4 + 2 = 6$$

$$2\alpha - \beta = 8 - 1 = 7$$

The equation with roots 6 and 7:

$$x^2 - (6+7)x + 6 \times 7 = 0$$

$$x^2 - 13x + 42 = 0$$

This matches option 3.