The correct options are
A 32
C 72
Given : |z−2|=min{|z−1|,|z−5|}
Let z=x+iy
When |z−1|<|z−5|
⇒(x−1)2+y2<(x−5)2+y2⇒−2x+1<−10x+25⇒x<3
Now,
|z−2|=|z−1|
z is perpendicular bisector of line segment joining points (2,0) and (1,0)
x=32
When |z−1|≥|z−5|
⇒x≥3
Now,
|z−2|=|z−5|
z is perpendicular bisector of line segment joining the points (2,0) and (5,0)
x=72
Hence, the possible values of Re(z) are 32 and 72.