TS EAMCET 5 May 2018 Shift 2 Solved Paper

The equation of plane passing through three non collinear points(0,1,2),(1,0,-2) and (-2,1,0) is
|

x−0	y−1	z−2
1−0	0−1	−2−2
−2−0	1−1	0−2

| = 0
|

x	y−1	z−2
1	−1	−4
−2	0	−2

| = 0
2x+10y−10−2z+4=0
2x+10y−2z−6=0The equation of the plane passing through the point (1,2,3) is
a(x−1)+b(y−2)+c(z−3)=0
The plane is perpendicular to x+y+z=1.
Therefore,
a+b+c=0.
Also, the plane is perpendicular to 2x−3y+z=5
Therefore,
2a−3b+c=0
Thus,

a

1+3

=

b

2−1

=

c

−3−2

=λ
Therefore,
4λϕ(x−1)+λ(y−2) −5λ(z−3)=0
4x−4+y−2−5z+15=0
4x+y−5z+9=0
Hence,
cos‌θ=|

8+10+10

√108√16+25+1

|
=|

28

√108√16+25+1

|
=

14

9√14

=

√14

9