JEE Advanced 2020 Concept Recapitulation Test 2 Paper 1

We have max {|x−1|,|y−2|}=4.
If |x−1|≥|y−2| then |x−1|=4,
i.e. if (x+y−3)(x−y+1)≥0 then x=−3 or 5 ,
If |y−2|≥|x−1| then |y−2|=4 i.e. (x+y−3)(x−y+1)≤0, then y=−2 or 6 .
So, the locus of P bounds a square, the equation of whosesides are
x=−3,x=5,y=−2,y=6
Thus the area is (8)2=64.