We have,
Z=4z+2i2z−3i
⇒Z=4x+i(4y+2)2x+i(2y−3)
⇒Z=4x+i(4y+2)2x+i(2y−3)×4x−i(4y+2)4x−i(4y+2)
⇒Z=16x2+(4y+2)28x2+(4y+2)(2y−3)+i[4x(2y−3)−(8xy+4x)]
Since, Z is purely real, so 16x2+(4y+2)24x(2y−3)−(8xy+4x)=0
⇒−12x+8xy−(8xy+4x)=0
⇒x=0
And,
4x+i(4y+2)=0
⇒i(4y+2)=0
⇒y=−21
Hence this is the required option.