TS EAMCET 5 May 2018 Shift 1 Solved Paper

The equation of the required plane is given as,
a(x−2)+b(y+1) +c(z−3)=0 …… (1)
since, the plane is perpendicular to 3x−2y+z=9 then,
3a−2b+c=0 …… (2)
The required plane is also perpendicular to x+y+z=9
,a+b+c=0 …… (3)
From equation ( 2 ) and equation ( 3 ),

a

−2−1

=

b

1−3

=

c

3+2

Assume,

a

−3

=

b

−2

=

c

5

=λ
The equation (1) can be written as,
−3λ(x−2)−2λ(y+1) +5λ(z−3)=0
x+

2

3

y−

5

3

z+

11

3

=0
By comparing, ax+by+cz+d=0 then,
d=

11

3