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=|
∧
i
∧
j
∧
k
−1
1
−1
2
−1
2
| =(2−1)
∧
i
−(−2+2
∧
j
+(1−2
∧
k
=
∧
i
−
∧
k
=⟨1,0,−1⟩ Now, equation of the point (2,1,2) and the normal vector (1,0,−1) is 1(x−2+0(y−1)−1(z−2)=0 ⇒x−2−z+2=0 ⇒x−z=0 Comparing this with ax+by+cz+d=0, we have a=1,b=0,c=−1,d=0 ∴
