Key Concepts and Formulas

1. Collinearity of Vectors: Two non-zero vectors $\vec{x}$ and $\vec{y}$ are collinear if and only if $\vec{x} = k\vec{y}$ for some non-zero scalar $k$.
2. Linear Independence of Non-Collinear Vectors: If two vectors $\vec{x}$ and $\vec{y}$ are non-collinear, they are linearly independent. This implies that if $p\vec{x} + q\vec{y} = \overrightarrow{0}$, then $p=0$ and $q=0$.
3. Vector Equation Manipulation: Algebraic operations such as addition, subtraction, and scalar multiplication can be applied to vector equations.

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Step-by-Step Solution

We are given three non-zero vectors $\vec{a}, \vec{b}, \vec{c}$ such that $\vec{b}$ and $\vec{c}$ are non-collinear.

Step 1: Express the first collinearity condition mathematically. We are given that $\vec{a}+5 \vec{b}$ is collinear with $\vec{c}$. By the definition of collinearity, this means there exists a non-zero scalar, say $k_1$, such that: $\vec{a} + 5\vec{b} = k_1 \vec{c}$ Rearranging this equation to express $\vec{a}$ in terms of $\vec{b}$ and $\vec{c}$: $\vec{a} = k_1 \vec{c} - 5\vec{b} \quad (*)$

Step 2: Express the second collinearity condition mathematically. We are given that $\vec{b}+6 \vec{c}$ is collinear with $\vec{a}$. This means there exists a non-zero scalar, say $k_2$, such that: $\vec{b} + 6\vec{c} = k_2 \vec{a} \quad (**)$

Step 3: Substitute the expression for $\vec{a}$ from Step 1 into the equation from Step 2. Substitute the expression for $\vec{a}$ from equation $(*)$ into equation $(**)$: $\vec{b} + 6\vec{c} = k_2 (k_1 \vec{c} - 5\vec{b})$ Distribute $k_2$ on the right side: $\vec{b} + 6\vec{c} = k_1 k_2 \vec{c} - 5k_2 \vec{b}$ Now, rearrange the terms to group $\vec{b}$ and $\vec{c}$ on one side, setting the expression equal to the zero vector: $\vec{b} + 5k_2 \vec{b} + 6\vec{c} - k_1 k_2 \vec{c} = \overrightarrow{0}$ Combine the coefficients of $\vec{b}$ and $\vec{c}$: $(1 + 5k_2)\vec{b} + (6 - k_1 k_2)\vec{c} = \overrightarrow{0}$

Step 4: Apply the concept of linear independence to solve for $k_1$ and $k_2$. Since $\vec{b}$ and $\vec{c}$ are non-collinear, they are linearly independent. For the equation $(1 + 5k_2)\vec{b} + (6 - k_1 k_2)\vec{c} = \overrightarrow{0}$ to hold, the coefficients of $\vec{b}$ and $\vec{c}$ must both be zero. This gives us a system of two linear equations:

1. $1 + 5k_2 = 0$
2. $6 - k_1 k_2 = 0$

From equation (1), we solve for $k_2$: $5k_2 = -1 \implies k_2 = -\frac{1}{5}$ Now, substitute the value of $k_2$ into equation (2): $6 - k_1 \left(-\frac{1}{5}\right) = 0$ $6 + \frac{k_1}{5} = 0$ $\frac{k_1}{5} = -6 \implies k_1 = -30$

Step 5: Use the final given equation to find $\alpha$ and $\beta$. We are given the equation $\vec{a}+\alpha \vec{b}+\beta \vec{c}=\overrightarrow{0}$. From Step 1, we have $\vec{a} = k_1 \vec{c} - 5\vec{b}$. Substituting the value $k_1 = -30$, we get: $\vec{a} = -30\vec{c} - 5\vec{b}$ Now substitute this expression for $\vec{a}$ into the given equation: $(-5\vec{b} - 30\vec{c}) + \alpha \vec{b} + \beta \vec{c} = \overrightarrow{0}$ Group the terms involving $\vec{b}$ and $\vec{c}$: $(\alpha - 5)\vec{b} + (\beta - 30)\vec{c} = \overrightarrow{0}$

Step 6: Apply linear independence again to find $\alpha$ and $\beta$. Since $\vec{b}$ and $\vec{c}$ are non-collinear, their coefficients must be zero: $\alpha - 5 = 0 \implies \alpha = 5$ $\beta - 30 = 0 \implies \beta = 30$

Step 7: Calculate $\alpha+\beta$. Finally, we compute the sum $\alpha+\beta$: $\alpha + \beta = 5 + 30 = 35$

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Common Mistakes & Tips

• Misinterpreting Collinearity: Ensure that collinearity is correctly translated into a scalar multiple relationship. If $\vec{x}$ and $\vec{y}$ are collinear, $\vec{x}=k\vec{y}$ where $k \neq 0$.
• Incorrect Application of Linear Independence: Linear independence of $\vec{b}$ and $\vec{c}$ is the key. Remember that if $p\vec{b} + q\vec{c} = \overrightarrow{0}$ and $\vec{b}, \vec{c}$ are non-collinear, then $p=0$ and $q=0$.
• Algebraic Errors: Double-check all algebraic manipulations, especially when substituting and rearranging terms.

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Summary

The problem involves using the definition of collinearity to set up scalar multiple equations between the given vectors. By strategically substituting these equations, we can form a linear combination of the non-collinear vectors $\vec{b}$ and $\vec{c}$. The principle of linear independence is then applied to equate the coefficients of $\vec{b}$ and $\vec{c}$ to zero, allowing us to solve for unknown scalars. This process is repeated to find the values of $\alpha$ and $\beta$, which are then summed to find the final answer.

The final answer is $\boxed{35}$.