In an equilateral triangle PQR , let the vertex P be at (3, 5) and...

In an equilateral triangle, the orthocentre coincides with the centroid.

The centroid divides the altitude from the vertex to the opposite side in the ratio $2:1$.

Let $M$ be the foot of the perpendicular from $P(3, 5)$ to the line $QR$ given by $x + y - 4 = 0$. The coordinates of $M$ can be found using the formula:

$\dfrac{x - 3}{1} = \dfrac{y - 5}{1} = -\dfrac{3 + 5 - 4}{1^2 + 1^2}$

$\dfrac{x - 3}{1} = \dfrac{y - 5}{1} = -2$

This gives $x = 1$ and $y = 3$. Thus, $M$ is $(1, 3)$.

The orthocentre $H(\alpha, \beta)$ divides the segment $PM$ internally in the ratio $2:1$. Using the section formula:

$\alpha = \dfrac{2(1) + 1(3)}{2 + 1} = \dfrac{5}{3}$

$\beta = \dfrac{2(3) + 1(5)}{2 + 1} = \dfrac{11}{3}$

Adding $\alpha$ and $\beta$:

$\alpha + \beta = \dfrac{5}{3} + \dfrac{11}{3} = \dfrac{16}{3}$

Therefore, $9(\alpha + \beta) = 9 \left( \dfrac{16}{3} \right) = 48$

Answer: $48$