Answer
(a) $x=-0.5m$
(b) region 3
(c) $x=-4.5m$
Work Step by Step
(a) We know that
$V_A=\frac{1}{4\pi \epsilon_{\circ}}(\frac{2q}{1.5+x}-\frac{q}{1.5-x})$
$\implies 0=\frac{2q}{1.5-x}-\frac{q}{1.5+x}$
$\implies \frac{2}{1.5-x}=\frac{1}{1.5+x}$
This simplifies to:
$x=-0.5m$
(b) We know that the electrical potential also vanishes in one of the given regions. As the lower charge $-q$ is less than the charge $+2q$ by a factor $2$, therefore, there should be a point located close to $-q$ by a factor of $2$ and thus the region is $3$, that is, $-3.5m \gt x\gt -5m$
(c) We know that
$V_3=\frac{1}{4\pi \epsilon_{\circ}}(\frac{2q}{1.5-x}-\frac{q}{-1.5-x})$
$\implies 0=\frac{2q}{1.5-x}-\frac{q}{-1.5-x}$
This simplifies to:
$x=-4.5m$