What is the distance between the planesx−2y+z−1=0 and−3x+6y−3z+2=0?

[d] Given planes are x−2y+z=1                             (i) And −3x+6y−3z=−2 ≡x−2y+z=32                             (ii) Since both planes are parallel and a=1,b=−2,c=1 and d1=−1,d2=−23 ∴ Distance =∣∣∣d2−d1a2+b2+c2−−−−−−−−−−√∣∣∣ Distance =∣∣∣1−231+4+1−−−−−−−√∣∣∣=136√