Let P1=(2,1,2) and P2=(1,2,1) So, P1P2=P2−P1 ‌⇒‌‌(1,2,1)−(2,1,2) ‌⇒‌‌(1−2,2−1,1−2 ‌⇒‌‌(−1,1,−1) The plane 2x−y+2z=1 has a normal vector n1=(2,−1,2). Since, the required plane is perpendicular to 2x−y+2z=1, its normal vector n=(a,b,c) must be perpendicular to n1. So, n must be perpendicular to P1P2. ∴n is parallel to the cross product of P1P2 and n1. n‌=P1P2×n1=|