The correct option is A c,b,a
I. x2+y2−6x−8y+12=0 and x2+y2−4x+6y+k=0
4≡(3,4)radius r1=√13
If this given circles cuts orthogonally then
2g1g2+2f1f2=c1+c2
So, 20(−3)(−2)+2(−6)(3)=12+k
12−24=12+k
k=−24
II. x2+y2−2x+3y+k=0 and x2+y2+8x−6y−7=0 cuts each othor othogonally than
2g1g2+2f1f2=c1+c2
So, 2x(−1)(4)+2(1/2)(−3)=k−7
−8−9=k−7
k=−10
III x2+y2+2x−2y+4=0 and x2+y2+4x−2y+2=0 are orthogonal to each other , so
2x(1)(2)+2x(−1)(−1)=4+2
6=6
True