The correct option is A (−2,0,−12)
P1≡x−2y−2z+1=0, P2≡2x−3y−6z+1=0
Pair of bisectors be
x−2y−2z+13=±2x−3y−6z+17
As a1a2+b1b2+c1c2=1(2)+(−2)(−3)+(−2)(−6)>0
−ve sign gives acute angle bisector
i.e., 7(x−2y−2z+1)=−3(2x−3y−6z+1)
⇒13x−23y−32z+10=0
Clearly (−2,0,−12) satisfy above plane.