For real numbers α and β ≠ 0, if the point of intersection of the straight lines (x-α)/1=(y-1)/2=(z-1)/3 and (x-4)/(β)=(y-6)/3=(z-7)/3, lies on the…

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