To find the orthocentre of a triangle, we need to find the point where the altitudes of the triangle intersect.

Given triangle sides:
x = 3
y = 4
3x - 4y = 6

Step 1: Convert the given equation into slope-intercept form:
3x - 4y = 6
-4y = -3x + 6
y = (3/4)x - 3/2

Step 2: Find the slopes of the sides of the triangle:
The slope of the side x = 3 is 0 (horizontal line).
The slope of the side y = 4 is undefined (vertical line).
The slope of the line y = (3/4)x - 3/2 is (3/4).

Step 3: Find the equations of the altitudes:
The altitude of the side x = 3 is a vertical line passing through the point (3, 0).
The equation of this altitude is x = 3.

The altitude of the side y = 4 is a horizontal line passing through the point (0, 4).
The equation of this altitude is y = 4.

The altitude of the side y = (3/4)x - 3/2 will be perpendicular to it and pass through the opposite vertex.
To find the equation of this altitude, we need to find the slope of the line perpendicular to (3/4).
The slope of the perpendicular line is the negative reciprocal of (3/4), which is -4/3.

Using the point-slope form of a line, we can find the equation of the altitude passing through the point (3, 4):
y - 4 = (-4/3)(x - 3)
3y - 12 = -4x + 12
4x + 3y = 24

Step 4: Find the point of intersection of the altitudes:
Now we need to solve the system of equations:
x = 3
y = 4
4x + 3y = 24

Solving these equations, we get:
x = 3
y = 4
4(3) + 3(4) = 24
12 + 12 = 24

Therefore, the point of intersection of the altitudes (orthocentre) is (3, 4).

Hence, the correct answer is option D: (3, 4).