Question

The complete set of values of $\alpha$ for which the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-3}{2}=\frac{y-5}{\alpha }=\frac{z-7}{\alpha +2}$ are coplanar is

• A. $\left\{2,3\right\}$
• B. $\left\{0,3\right\}$
• C. $\left[-2,3\right]$
• D. $R$

Answer 1

Answer: (D)

Any point on the first line is $\left(1+2\lambda ,2+3\lambda ,3+4\lambda ,\right)$
Any point on the second line is $\left(3+2\mu ,5+\alpha \mu ,7+2\mu +\alpha \mu \right)$
Now, as they are intersecting $1+2\lambda =3+2\mu$ and $2+3\lambda =5+\alpha \mu$
So, $\lambda =\mu +1,$ hence $2+3\left(\mu +1\right)=5+\alpha \mu$
$\Rightarrow \left(\alpha -3\right)\mu =0$
Hence for $\lambda =1,\mu =0,$ the point $\left(3,5,7\right)$ lies on both the lines irrespective of $\alpha .$
Hence, $\forall \alpha \in R$ lines are concurrent.