BITSAT 2019 Slot 2 Solved Paper

Equation of plane containing line x−y−1=0 and z−1=0 is (x−y−1)+λ(z−1)=0
⇒x−y+λz−1−λ=0 ..........(i)
Since, this plane is parallel to line

x

2

−

z

3

=1 and y=3

x−1

2

=

y−3

0

=

z

3

..........(ii)
∴2(1)+(0)(−1)+3(λ)=0
⇒ λ=−2∕3
Putting the value of lambda in Eq. (i), we get
x−y−

2

3

z−1+

2

3

=0; 3x−3y−2z=1