For real numbers $\alpha$ and $\beta \neq 0$, 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 $x+2 y-z=8$, then $\alpha-\beta$ is equal to:
Correct Option: , 4
First line is $(\phi+\alpha, 2 \phi+1,3 \phi+1)$
and second line is $(\mathrm{q} \beta+4,3 \mathrm{q}+6,3 \mathrm{q}+7)$.
For intersection $\phi+\alpha=\mathrm{q} \beta+4$ ............(I)
$2 \phi+1=3 q+6$ .................(I)
$3 \phi+1=3 q+7$ ................(III)
for (ii) \& (iii) $\phi=1, \mathrm{q}=-1$
So, from (i) $\alpha+\beta=3$
Now, point of intersection is $(\alpha+1,3,4)$
It lies on the plane.
Hence, $\alpha=5 \& \beta=-2$