Explanation

(d)
$\left| 3\mathcal{z -}1 \right| = 3\left| \mathcal{z -}2 \right|$
$\Rightarrow \ \ \left| \mathcal{z -}\frac{1}{3} \right|\mathcal{= |z -}2|$
$\mathcal{\Rightarrow \ z}$ is perpendicular bisector of $\left( \frac{1}{3}\ ,0 \right)\ and\ (2,0)$
$\Rightarrow \ \ x = \frac{7}{6}$