The equation of plane passing through (1,−1,2) is a(x−1)+b(y+1)+c(z−2)=0 Since plane (i) is perpendicular to the planes x+2y−2z=4 and ‌3x+2y+z=6 ‌∴‌‌a+2b−2c=0 ‌‌ and ‌3a+2b+c=0 ‌⇒‌
a
6
=‌
b
−7
=‌
c
−4
∴‌‌ The equation of the required plane is ‌6(x−1)−7(y+1)−4(z−2)=0 ‌⇒6x−7y−4z−5=0