Answer
There are no pairs with a point to the right of 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):
Choose any point to the right of the two particles.
Let $d_1$ be the distance between the point and the $-2q$ charge.
Let $d_2$ be the distance between the point and the $+6q$ charge.
Note that $d_1 \gt d_2$
We can find an expression for $V_{net}$ at this point:
$V_{net} = \frac{1}{4\pi~\epsilon_0}\sum \frac{q_i}{r_i}$
$V_{net} = \frac{1}{4\pi~\epsilon_0}(\frac{-2q}{d_1}+\frac{6q}{d_2})$
$V_{net} \gt 0$
Arrangement (2):
Choose any point to the right of the two particles.
Let $d_1$ be the distance between the point and the $+3q$ charge.
Let $d_2$ be the distance between the point and the $-4q$ charge.
Note that $d_1 \gt d_2$
We can find an expression for $V_{net}$ at this point:
$V_{net} = \frac{1}{4\pi~\epsilon_0}\sum \frac{q_i}{r_i}$
$V_{net} = \frac{1}{4\pi~\epsilon_0}(\frac{+3q}{d_1}-\frac{4q}{d_2})$
$V_{net} \lt 0$
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:
There are no pairs with a point to the right of the particles at which $V_{net} = 0$
