KCET 2019 Math Solved Paper

(C)
The dr’s of PA are x1 – 1, y1 – 3, z1 – 4
The dr’s n are 2, -1, 1
The dr’s of PA and n are parallel
∴

x1−1

2

=

y1−3

−1

=

z1−4

1

=λ
x1=2λ+1 , y1=−λ+3 , z1=λ+4
A = (2λ+1,−λ+3,λ+4) lies on 2 x - y + z + 3 = 0
⇒ −4λ+2−λ−3 - λ+4+3=0
⇒ 6λ = -6
⇒ λ=−1
∴ P = (–1, 4, 3)