1. Let $\vec a+5\vec b \text{ is collinear with } \vec c.$ So there exists a scalar $\lambda$ such that \vec a+5\vec b=\lambda \vec c \quad \Rightarrow \quad \vec a=-5\vec b+\lambda \vec c. \tag{1}

2. Also, $\vec b+6\vec c \text{ is collinear with } \vec a.$ Hence there exists a scalar $\mu$ such that \vec b+6\vec c=\mu \vec a. \tag{2}

3. Substitute (1) into (2): $\vec b+6\vec c=\mu(-5\vec b+\lambda \vec c).$ Comparing coefficients of the non-collinear vectors $\vec b$ and $\vec c$: $1=-5\mu, \qquad 6=\mu\lambda.$

   From $1=-5\mu$, $\mu=-\frac{1}{5}.$ Then $6=\left(-\frac15\right)\lambda \Rightarrow \lambda=-30.$

4. Therefore from (1), $\vec a=-5\vec b-30\vec c.$

5. Given $\vec a+\alpha \vec b+\beta \vec c=\vec 0.$ Substituting $\vec a$: $(-5\vec b-30\vec c)+\alpha \vec b+\beta \vec c=\vec 0,$ $(\alpha-5)\vec b+(\beta-30)\vec c=\vec 0.$

   Since $\vec b$ and $\vec c$ are non-collinear, they are linearly independent. Hence $\alpha-5=0, \qquad \beta-30=0.$ So $\alpha=5, \qquad \beta=30.$

6. Thus, $\alpha+\beta=5+30=35.$

Therefore, the correct option is D.