If $\frac{Z-2}{Z-6i}$  is purely real then the locus of $Z$  is ( where z=x+iy)

$\frac{\left(x-2\right)+iy}{x+i\left(y-6\right)}$  is purely real

I.P = 0 $\Rightarrow$  xy – (x – 2) (y – 6) = 0

           $\Rightarrow$  xy – [xy – 6x – 2y + 12] = 0        $\Rightarrow$ 6x + 2y – 12 = 0  $\Rightarrow$  3x + y – 6= 0