If the orthocentre of the triangle, whose vertices are $$1, 2$$, $$2, 3$$ and $$3, 1$$ 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):

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

Now finding the roots of the quadratic:

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

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

The quadratic equation is: