The correct option is
B P3<P2<P1Given
xy+x+y+1=0
comparing above eq with ax2+2hxy+by2+2gx+2fy+c=0
we get h=12,g=12,f=12,c=1
Product of perpendicular from origin is ∣∣
∣∣c√(a−b)2+4h2∣∣
∣∣
P1=∣∣
∣
∣
∣
∣
∣∣1√4(12)2∣∣
∣
∣
∣
∣
∣∣
P1=1√4(14)
P1=1√1
P1=1
Now given eq x2−y2+2x+1=0
comparing above eq with ax2+2hxy+by2+2gx+2fy+c=0
we get a=1,b=−1,h=0,g=1,f=0,c=1
Product of perpendicular from origin is ∣∣
∣∣c√(a−b)2+4h2∣∣
∣∣
P2=∣∣
∣
∣∣1√(1+1)2∣∣
∣
∣∣
P2=1√4
P2=12
Now given eq 2x2+3xy−2y2+3x+y+1=0
comparing above eq with ax2+2hxy+by2+2gx+2fy+c=0
we get a=2,b=−2,h=32,g=32,f=12,c=1
Product of perpendicular from origin is ∣∣
∣∣c√(a−b)2+4h2∣∣
∣∣
P3=∣∣
∣
∣
∣
∣
∣∣1√(2+2)2−4(32)2∣∣
∣
∣
∣
∣
∣∣
P3=∣∣
∣
∣
∣
∣
∣∣1√16−4(94)∣∣
∣
∣
∣
∣
∣∣
P3=∣∣∣1√16−9∣∣∣
P3=1√7
SO P3<P2<P1