Let $P$ be $a$ plane containing the line $\frac{x-1}{3}=\frac{y+6}{4}=\frac{z+5}{2}$ and parallel to the line $\frac{x-3}{4}=\frac{y-2}{-3}=\frac{z+5}{7}$. If the point $(1,-1, \alpha)$ lies on the plane P , then the value of $|5 \alpha|$ is equal to $\_\_\_\_$ .
