Consider the set of all lines px + qy + r = 0 such that 3p + 2q + 4r = 0. Which one of the following statements is true?

Given the equation of the lines as px + qy + r = 0, and the condition 3p + 2q + 4r = 0.
Let's analyze the condition 3p + 2q + 4r = 0. We can rewrite it as:
3p + 2q = -4r
p(3/4r) + q(2/4r) = -1
p(3/4r) + q(1/2r) = -1
If we consider the equation of the line px + qy + r = 0 and divide it by r, we get (p/r)x + (q/r)y + 1 = 0
Substituting the values from 3p + 2q + 4r = 0, we have
3p + 2q = -4r
(3p/4r) + (2q/4r) = -1
(3p/4r) + (q/2r) = -1
Let's assume r is not equal to 0. Dividing the line equation by -r, we have: -(p/r)x - (q/r)y = 1
Let's consider the condition 3p + 2q + 4r = 0. If we divide by -r, we get:
-3(p/r) - 2(q/r) = 4
Let's rearrange the line equation as: (p/r)x + (q/r)y = -1
From the condition 3p + 2q + 4r = 0, we can solve for r: r = -(3p + 2q)/4
Substitute this into the line equation:
px + qy - (3p + 2q)/4 = 0
4px + 4qy - 3p - 2q = 0
p(4x - 3) + q(4y - 2) = 0
For this equation to hold for all p and q, we must have:
4x - 3 = 0 => x = 3/4
4y - 2 = 0 => y = 1/2
Therefore, all lines pass through the point (3/4, 1/2).
Hence, the correct statement is that all points pass through (3/4, 1/2).