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