AP EAMCET Engineering 20 Apr 2019 Shift 1 Paper

The equation of the plane passes through the point
(i+2j−k)=(1,2,−1) having direction ratios normal to a,b,c is,
a(x−1)+b(y−2)+c(z+1)=0‌‌‌..........(I)
The plane in equation (I) is perpendicular to line of intersection
of planes r⋅(3i−j+k)=1 and r⋅(i+4j−2k)=2so
3a−b+c=0
a+4b−2c=0
This implies,
‌

a

2−4

‌‌=‌

−b

−6−1

=‌

c

12+1

‌

a

2

‌‌=‌

b

−7

=‌

c

−13

So, equation of required plane is
2(x−1)−7(y−2)−13(z+1)=0
2x−7y−13z−2+14−13‌‌=0
2x−7y−13z‌‌=1
r⋅(2i−7j−13k)‌‌=1