For real numbers α and β ≠ 0 , if the point of intersection of the straight lines x - α…

Given, $\frac{x-\alpha }{1}=\frac{y-1}{2}=\frac{z-1}{3}$ and $\frac{x-4}{\beta }=\frac{y-6}{3}=\frac{z-7}{3}$ lies on the plane $x+2y-z=8$.

Let $\frac{x-\alpha }{1}=\frac{y-1}{2}=\frac{z-1}{3}=\varnothing$

and $\frac{x-4}{\beta }=\frac{y-6}{3}=\frac{z-7}{3}=q$

Any point on the first line can be considered as $\left(\varphi +\alpha , 2\varphi +1, 3\varphi +1\right)$

and a point on the second line can be $\left(q\beta +4,3q+6,3q+7\right).$

For intersection of the lines $\varphi +\alpha =q\beta +4 ...\left(1\right)$

$2\varphi +1=3q+6 ...\left(2\right)$

$3\varphi +1=3q+7 ...\left(3\right)$

For $\left(2\right) \& \left(3\right)$ $\varphi =1,q=-1$

So, from $\left(1\right)$ $\alpha +\beta =3$

Now, point of intersection is $\left(\alpha +1,3,4\right)$

It lies on the given plane, therefore, $\alpha +1+2\times 3-4=8\Rightarrow \alpha =5$

 $\Rightarrow \beta =-2$ from $\alpha +\beta =3$

Hence, $\alpha -\beta =5-\left(-2\right)=7$