Explanation

(a)
Let $z = x + iy$
$\therefore\ \ \ |z - 1| = |z - 2| = |z - i|$
$\Rightarrow \ \ \ \ \ \ \left| (x - 1) \pm iy \right| = \left| (x - 2) + iy \right|$
$= \left| x + i(y - 1) \right|\ \ \ \ \ \ \ \ \lbrack$put$\ z = x + iy\rbrack$
$\Rightarrow \ \ \ \ x^{2} - 2x + 1 + y^{2} = x^{2} + 4 - 4x + y^{2}$
$= x^{2} + y^{2} + 1 - 2y$
Taking Ist and IInd terms
$- 2x + 1 = 4 - 4x\ \ \ \Rightarrow \ \ \ 2x = 3$ …(i)
Taking IInd and IIIrd terms
$4 - 4x = 1 - 2y\ \ \ \Rightarrow \ \ \ 4x - 2y = 3$ …(ii)
Taking Ist and IIIrd terms
$- 2x + 1 = 1 - 2y\ \ \ \ \Rightarrow \ \ \ x = y$ …(iii)
From Eq. (i), $x = \frac{3}{2}$
From Eqs. (i) and (iii), $y = \frac{3}{2}$
On putting the value of $x$ and $y$ in Eq. (ii), we get
$4\left( \frac{3}{2} \right) - 2\left( \frac{3}{2} \right) = 3\ \ \ \ \ \Rightarrow \ \ \ 3 = 3$
$\therefore$ One solution exists.