If the orthocentre of the triangle, whose vertices are $$\left(1,2\right)$$, $$\left(2,3\right)$$ and $$\left(3,1\right)$$ is $$\alpha, \beta$$, then the quadratic equation whose roots are $$\alpha + 4\beta$$ and $$4\alpha + \beta$$, is

Given vertices: $$A(1, 2)$$, $$B(2, 3)$$, $$C(3, 1)$$.

To find the orthocenter, we need the intersection of any two altitudes.

Altitude from C perpendicular to AB:

Slope of AB = $$\frac{3-2}{2-1} = 1$$. So slope of altitude from C = $$-1$$.

Equation: $$y - 1 = -1(x - 3) \Rightarrow y = -x + 4$$ ... (i)

Altitude from A perpendicular to BC:

Slope of BC = $$\frac{1-3}{3-2} = -2$$. So slope of altitude from A = $$\frac{1}{2}$$.

Equation: $$y - 2 = \frac{1}{2}(x - 1) \Rightarrow y = \frac{x}{2} + \frac{3}{2}$$ ... (ii)

Solving (i) and (ii):

$$ -x + 4 = \frac{x}{2} + \frac{3}{2} $$

$$ -\frac{3x}{2} = -\frac{5}{2} $$

$$ x = \frac{5}{3}, \quad y = -\frac{5}{3} + 4 = \frac{7}{3} $$

So the orthocenter is $$(\alpha, \beta) = \left(\frac{5}{3}, \frac{7}{3}\right)$$.

Now finding the roots of the quadratic:

$$ \alpha + 4\beta = \frac{5}{3} + \frac{28}{3} = \frac{33}{3} = 11 $$

$$ 4\alpha + \beta = \frac{20}{3} + \frac{7}{3} = \frac{27}{3} = 9 $$

Sum of roots = $$11 + 9 = 20$$

Product of roots = $$11 \times 9 = 99$$

The quadratic equation is:

$$ x^2 - 20x + 99 = 0 $$