3d Geometry

Let the image of the point $\mathrm{P}(1,6, a)$ in the line $\mathrm{L}: \frac{x}{1}=\frac{y-1}{2}=\frac{z-a+1}{b}, b>0$, be $\left(\frac{a}{3}, 0, a+c\right)$. If $\mathrm{S}(\alpha, \beta, \gamma), \alpha>0$, is the point on L such that the distance of S from the foot of perpendicular from the point P on L is $2 \sqrt{14}$, then $\alpha+\beta+\gamma$ is equal to: