The correct option is
B 92The co-ordinates of any point on the first line are given by
x−12=y+13=z−14=λ
⇒x−1=2λ,y+1=3λ,z−1=4λ
⇒x=1+2λ,y=1+3λ,z=1+4λ
Thus, the co-ordiantes of any point on this line are (1+2λ,1+3λ,1+4λ)
The co-ordiantes of any point on the second line are
x−31=y−k2=z1=μ
⇒x−3=μ,y−k=2μ,z=μ
⇒x=3+μ,y=k+2μ,z=μ
Thus, the co-ordiantes of any point on second line are (3+μ,k+μ,μ)
If these two lines intersect each other, then
3+μ=1+2λ ,k+2μ=1+3λ and μ=1+4λ
⇒2λ−μ=2, 3λ−2μ=k+1 and 4λ−μ=−1
Solving the equations 2λ−μ=2 and 4λ−μ=−1
⇒2λ−μ−4λ+μ=2−(−1)
⇒−2λ=2+1=3 or λ=−32
Substituting the value of λ=−32 in 2λ−μ=2
we get 2×−32−μ=2 or −3−μ=2 or μ=−5
Substituting the values of λ=−32 and μ=−5 in 3λ−2μ=k+1
we get
3×−32−2×−5=k+1
k+1=−92+10=−9+202=112
∴k+1=112
⇒k=112−1=11−22=92