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+5vec b` is collinear with `vec c,vec b+6vec c` is collinear with `vec a` and `vec a+alpha vec b+betavec c=vec 0` then `alpha+beta` is equal to

Options

A. 30
B. 35
C. -30
D. -25

Answer

To solve the problem, we need to analyze the given conditions step by step.

### Step 1: Understanding Collinearity Conditions
We are given that:
1. \(\vec{a} + 5\vec{b}\) is collinear with \(\vec{c}\).
2. \(\vec{b} + 6\vec{c}\) is collinear with \(\vec{a}\).

From the first condition, we can express it as:
\[ \vec{a} + 5\vec{b} = \lambda \vec{c} \quad \text{(for some scalar } \lambda\text{)} \]
From this, we can rearrange to find \(\vec{a}\):
\[ \vec{a} = \lambda \vec{c} - 5\vec{b} \tag{1} \]

From the second condition, we can express it as:
\[ \vec{b} + 6\vec{c} = \mu \vec{a} \quad \text{(for some scalar } \mu\text{)} \]
Rearranging gives us:
\[ \vec{a} = \frac{1}{\mu} \vec{b} + \frac{6}{\mu} \vec{c} \tag{2} \]

### Step 2: Equating the Two Expressions for \(\vec{a}\)
Now we have two expressions for \(\vec{a}\) from equations (1) and (2):
\[ \lambda \vec{c} - 5\vec{b} = \frac{1}{\mu} \vec{b} + \frac{6}{\mu} \vec{c} \]

### Step 3: Rearranging and Collecting Terms
Rearranging gives us:
\[ \lambda \vec{c} - \frac{6}{\mu} \vec{c} = 5\vec{b} + \frac{1}{\mu} \vec{b} \]
Factoring out \(\vec{c}\) and \(\vec{b}\):
\[ \left(\lambda - \frac{6}{\mu}\right) \vec{c} = \left(5 + \frac{1}{\mu}\right) \vec{b} \]

### Step 4: Coefficients Must Be Equal
Since \(\vec{b}\) and \(\vec{c}\) are non-collinear, the coefficients must be equal:
1. \(\lambda - \frac{6}{\mu} = 0\)
2. \(5 + \frac{1}{\mu} = 0\)

### Step 5: Solving the Equations
From the second equation:
\[ \frac{1}{\mu} = -5 \implies \mu = -\frac{1}{5} \]
Substituting \(\mu\) into the first equation:
\[ \lambda - \frac{6}{-\frac{1}{5}} = 0 \implies \lambda + 30 = 0 \implies \lambda = -30 \]

### Step 6: Using the Third Condition
Now we use the third condition:
\[ \vec{a} + \alpha \vec{b} + \beta \vec{c} = \vec{0} \]
Substituting \(\vec{a}\) from equation (1):
\[ \lambda \vec{c} - 5\vec{b} + \alpha \vec{b} + \beta \vec{c} = \vec{0} \]
This simplifies to:
\[ (\lambda + \beta) \vec{c} + (-5 + \alpha) \vec{b} = \vec{0} \]
For this to hold, the coefficients must be zero:
1. \(\lambda + \beta = 0\)
2. \(-5 + \alpha = 0\)

### Step 7: Solving for \(\alpha\) and \(\beta\)
From \(-5 + \alpha = 0\):
\[ \alpha = 5 \]
From \(\lambda + \beta = 0\):
\[ -30 + \beta = 0 \implies \beta = 30 \]

### Step 8: Finding \(\alpha + \beta\)
Now we can find:
\[ \alpha + \beta = 5 + 30 = 35 \]

### Final Answer
Thus, the value of \(\alpha + \beta\) is:
\[ \boxed{35} \]