Equation of plane passing through (−1,1,1) is a(x+1)+b(y−1)+c(z−1)=0 . . . (i) Also, it is passing through (1,−1,1). ∴‌a(1+1)+b(−1−1)+c(1−1)=0 ⇒2a−2b+0c=0 . . . (ii) Also, required equation of plane (i) is perpendicular to x+2y+2z=5. ∴‌a×1+b×2+c×2‌=0 ‌⇒a+2b+2c‌=0 . . . (iii) Eqs. (ii) and (iii) are identical. ∴‌‌
a
−4−0
=‌
−b
4−0
=‌
c
4+2
⇒‌
a
−4
=‌
b
−4
=‌
c
6
⇒‌
a
−2
=‌
b
−2
=‌
c
3
=λ (say) ⇒a=−2λ,b=−2λ,c=3λ On putting the values of a,b and c in Eq. (i), we get ‌−2λ(x+1)−2λ(y−1)+3λ(z−1)=0 ‌⇒‌‌λ[−2x−2−2y+2+3z−3]=0 ‌⇒‌‌−2x−2y+3z−3=0 ‌⇒‌‌2x+2y−3z+3=0 ‌