Given planes are perpendicular to the plane ( π ) ∴ Normal vector to the plane ( π ) is given by |
∧
i
∧
j
∧
k
1
1
−1
2
−1
1
|&=0
∧
i
−3
∧
j
−3
∧
k
=−3(0
∧
i
+
∧
j
+
∧
k
) Direction Ratio's of −3(0
∧
i
+
∧
j
+
∧
k
) is (0,1,1) Plane ( π ) is passing through the point ( 1,2,−3 ) ∴ Plane (π)0(x−1)+1(y−2)+1(z+3)=0 ⇒y−2+z+3=0 ⇒0x+y+z+1=0 Given Plane (π)ax+by+cz+1=0 ∴a=0,b=1 and c=1 Now, a2+b2+c2=02+12+12 =0+1+1=2
