Question

For real numbers $\alpha$ and $\beta \ne 0$, if the point of intersection of the straight lines $\frac{x-\alpha }{1}=\frac{y-1}{2}=\frac{z-1}{3}$ and $\frac{x-4}{\beta }=\frac{y-6}{3}=\frac{z-7}{3}$, lies on the plane $x+2y-z=8,$ then $\alpha –\beta$ is equal to

• A. $9$
• B. $5$
• C. $3$
• D. $7$

Answer 1

The correct option is D $7$

Let $L_1:\frac{x-\alpha }{1}=\frac{y-1}{2}=\frac{z-1}{3}=\lambda$ and $L_2:\frac{x-4}{\beta }=\frac{y-6}{3}=\frac{z-7}{3}=\mu$
Let point on line $L_1$ be $(\lambda +\alpha ,2\lambda +1,3\lambda +1)$ and a point on line $L_2$ be $(\mu \beta +4,3\mu +6,3\mu +7)$
For point of intersection
$\lambda +\alpha =\mu \beta +4,2\lambda +1=3\mu +6\&3\lambda +1=3\mu +7$
$\lambda =1$ and $\mu =-1$
$\Rightarrow 1+\alpha =-\beta +4\Rightarrow \alpha +\beta =3$
$\therefore$ Point of intersection $(1+\alpha ,3,4)$
Above point lies on the plane $x+2y-z=8,$
$\Rightarrow 1+\alpha +6-4=8\Rightarrow \alpha =5,\beta =-2$
We get, $\alpha -\beta =7$