Answer
In Arrangement (1) and Arrangement (2), there is a point between the particles at which $V_{net} = 0$
Work Step by Step
We can write a general equation for the net electric potential:
$V_{net} = \frac{1}{4\pi~\epsilon_0}\sum \frac{q_i}{r_i}$
Arrangement (1):
Let $d$ be the distance between the two charges.
Consider the point a distance of $\frac{d}{4}$ to the right of the $-2q$ charge.
We can find $V_{net}$ at this point:
$V_{net} = \frac{1}{4\pi~\epsilon_0}\sum \frac{q_i}{r_i}$
$V_{net} = \frac{1}{4\pi~\epsilon_0}(\frac{-2q}{d/4}+\frac{6q}{3d/4})$
$V_{net} = \frac{1}{4\pi~\epsilon_0}(\frac{-8q}{d}+\frac{8q}{d})$
$V_{net} = 0$
Arrangement (2):
Let $d$ be the distance between the two charges.
Consider the point a distance of $\frac{3d}{7}$ to the right of the $+3q$ charge.
We can find $V_{net}$ at this point:
$V_{net} = \frac{1}{4\pi~\epsilon_0}\sum \frac{q_i}{r_i}$
$V_{net} = \frac{1}{4\pi~\epsilon_0}(\frac{3q}{3d/7}+\frac{-4q}{4d/7})$
$V_{net} = \frac{1}{4\pi~\epsilon_0}(\frac{7q}{d}-\frac{7q}{d})$
$V_{net} = 0$
In Arrangement (3), $V_{net} \gt 0$ at all points since both charges are positive.
In Arrangement (4), $V_{net} \lt 0$ at all points since both charges are negative.
Therefore:
In Arrangement (1) and Arrangement (2), there is a point between the particles at which $V_{net} = 0$
