MHT CET 2017 Math Solved Paper

(b) Since,

x−1

2

=

y+1

2

=

z−1

1

= λ ......(i)
and ,

x−3

1

=

y−6

2

=

z

1

......(ii)
Now, any point on the line (i) is P(2 λ + 1 , 2λ-1, 4λ+1)
∴

2λ+1−3

1

=

2λ−1−6

2

=

4λ+1

1

[from (ii)]
So, 4 λ - 4 = 2λ - 7 ⇒ 2λ = -3
Hence, point of intersection P is
=(2×(−

3

2

)+1 , 2×(−

3

2

)−1 , 4×(−

3

2

)+1)
≡(−2,−4,−5)