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

We need to find the quadratic equation whose roots are $$\alpha + 4\beta$$ and $$4\alpha + \beta$$, where $$(\alpha, \beta)$$ is the orthocentre of the triangle with vertices $$(1,2)$$, $$(2,3)$$, $$(3,1)$$.

Let $$A(1,2)$$, $$B(2,3)$$, $$C(3,1)$$.

Slope of $$BC = \frac{1-3}{3-2} = -2$$. The altitude from $$A$$ is perpendicular to $$BC$$: slope $$= \frac{1}{2}$$.

Altitude from $$A$$: $$y - 2 = \frac{1}{2}(x-1)$$, i.e., $$2y - 4 = x - 1$$, i.e., $$x - 2y = -3$$ ... (i)

Slope of $$AC = \frac{1-2}{3-1} = -\frac{1}{2}$$. The altitude from $$B$$ is perpendicular to $$AC$$: slope $$= 2$$.

Altitude from $$B$$: $$y - 3 = 2(x-2)$$, i.e., $$y = 2x - 1$$ ... (ii)

Substituting (ii) in (i): $$x - 2(2x-1) = -3 \Rightarrow x - 4x + 2 = -3 \Rightarrow -3x = -5 \Rightarrow x = \frac{5}{3}$$.

$$y = 2 \times \frac{5}{3} - 1 = \frac{7}{3}$$.

So $$\alpha = \frac{5}{3}$$, $$\beta = \frac{7}{3}$$.

Root 1: $$\alpha + 4\beta = \frac{5}{3} + \frac{28}{3} = \frac{33}{3} = 11$$.

Root 2: $$4\alpha + \beta = \frac{20}{3} + \frac{7}{3} = \frac{27}{3} = 9$$.

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

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

Equation: $$x^2 - 20x + 99 = 0$$.

This matches Option D: $$x^2 - 20x + 99 = 0$$.

Therefore, the answer is Option D.