Answer:
Option D
Explanation:
Given lines are ,
$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda_{1}$ (let) ..........(i)
and $\frac{x-3}{1}=\frac{y-\lambda}{2}=\frac{z}{1}=\lambda_{2}$ (let) ......(ii)
Then, any point on the lie (i) is
$(2\lambda_{1}+1, 3\lambda_{1}-1,4\lambda_{1}+1)$ and any point on line (ii) is
$(\lambda_{2}+3, 2\lambda_{2}+\lambda,\lambda_{2})$
Clearly, then lines (i) and (ii) will intersect , if
$(2\lambda_{1}+1, 3\lambda_{1}-1,4\lambda_{1}+1)$ = $(\lambda_{2}+3, 2\lambda_{2}+\lambda,\lambda_{2})$
For some particular value of $\lambda_{1}$ and $\lambda_{2}$
$\Rightarrow$ $ 2\lambda_{1}+1=\lambda_{2}+3, 3\lambda_{1}-1=2\lambda_{2}+\lambda, and 4\lambda_{1}+1=\lambda_{2}$
$\Rightarrow$ $ 2\lambda_{1}-\lambda_{2}=2,3\lambda_{1}-2\lambda_{2}=\lambda+1 $ and$ 4\lambda_{1}-\lambda_{2}=-1$
On solving $2\lambda_{1}-\lambda_{2}=2,4\lambda_{1}-\lambda_{2}=-1$
We get, $\lambda_{1}=-\frac{3}{2}$ and $\lambda_{2}=-5$
Now putting the values of $\lambda_{1}$ and $\lambda_{2}$ in
$3\lambda_{1}-2\lambda_{2}=\lambda+1$
$\Rightarrow$ $ 3(\frac{-3}{2})-2(-5)=\lambda+1\Rightarrow \frac{-9}{2}+10=\lambda+1$
$\Rightarrow $ $\lambda+1=\frac{11}{2}\Rightarrow\lambda=\frac{9}{2}$
