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

We have the family of lines

$$p\,x+q\,y+r=0$$

subject to the linear condition

$$3p+2q+4r=0.$$

From the second relation we can express r in terms of p and q. Solving for r,

$$4r=-(3p+2q)\;\Longrightarrow\; r=-\dfrac{3p+2q}{4}.$$

Substituting this value of r back into the equation of the line,

$$p\,x+q\,y-\dfrac{3p+2q}{4}=0.$$

To clear the fraction, multiply every term by 4:

$$4p\,x+4q\,y-3p-2q=0.$$

Group the terms containing p and those containing q:

$$p\,(4x-3)+q\,(4y-2)=0.$$

In this expression p and q are arbitrary real numbers, not both zero. For the left-hand side to vanish for every possible choice of (p,q) that satisfies the original constraint, each of the factors that multiplies p and q must itself be zero. Thus we require

$$4x-3=0 \quad\text{and}\quad 4y-2=0.$$

Solving these two simple linear equations gives

$$x=\dfrac34, \qquad y=\dfrac12.$$

So every line in the given set passes through the single fixed point $$\left(\dfrac34,\dfrac12\right).$$

Because all the lines meet at one common point, they are concurrent, and the point of concurrency is exactly $$\left(\dfrac34,\dfrac12\right).$$

Hence, the correct answer is Option B.