Let $$\vec{a}, \vec{b}$$ and $$\vec{c}$$ be three non-zero vectors such that $$\vec{b}$$ and $$\vec{c}$$ are non-collinear. If $$\vec{a} + 5\vec{b}$$ is collinear with $$\vec{c}$$, $$\vec{b} + 6\vec{c}$$ is collinear with $$\vec{a}$$ and $$\vec{a} + \alpha\vec{b} + \beta\vec{c} = \vec{0}$$, then $$\alpha + \beta$$ is equal to

$$\vec{a} + 5\vec{b}$$ is collinear with $$\vec{c}$$, and $$\vec{b} + 6\vec{c}$$ is collinear with $$\vec{a}$$. Also $$\vec{a} + \alpha\vec{b} + \beta\vec{c} = \vec{0}$$.

Express the collinearity conditions.

Since $$\vec{a} + 5\vec{b}$$ is collinear with $$\vec{c}$$:

$$\vec{a} + 5\vec{b} = \lambda\vec{c}$$ for some scalar $$\lambda$$ ... (i)

Since $$\vec{b} + 6\vec{c}$$ is collinear with $$\vec{a}$$:

$$\vec{b} + 6\vec{c} = \mu\vec{a}$$ for some scalar $$\mu$$ ... (ii)

Express $$\vec{a}$$ from equation (ii).

$$\vec{a} = \frac{1}{\mu}\vec{b} + \frac{6}{\mu}\vec{c}$$

Substitute into equation (i).

$$\frac{1}{\mu}\vec{b} + \frac{6}{\mu}\vec{c} + 5\vec{b} = \lambda\vec{c}$$

$$\left(\frac{1}{\mu} + 5\right)\vec{b} + \left(\frac{6}{\mu} - \lambda\right)\vec{c} = \vec{0}$$

Since $$\vec{b}$$ and $$\vec{c}$$ are non-collinear, both coefficients must be zero:

$$\frac{1}{\mu} + 5 = 0 \implies \mu = -\frac{1}{5}$$

$$\frac{6}{\mu} = \lambda \implies \lambda = 6 \times (-5) = -30$$

Find $$\alpha$$ and $$\beta$$.

From (i): $$\vec{a} = -5\vec{b} + \lambda\vec{c} = -5\vec{b} - 30\vec{c}$$

So $$\vec{a} + 5\vec{b} + 30\vec{c} = \vec{0}$$

Therefore $$\alpha = 5$$ and $$\beta = 30$$.

$$\alpha + \beta = 5 + 30 = 35$$

The answer is Option 1: 35.