Given `veca, vecb and vecc` are 3 vectors such that `vecb, vecc` are parallel unit vectors and `|veca|=3`. If `veca+lambdavecc=4vecb`, then the sum of all the possible positive values of `lambda` is

Answer

To solve the given problem step by step, we will follow the logical deductions based on the properties of vectors and algebraic manipulations.

### Step 1: Understand the given information
We have three vectors: \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). It is given that:
- \(\vec{b}\) and \(\vec{c}\) are parallel unit vectors.
- \(|\vec{a}| = 3\).
- The equation \(\vec{a} + \lambda \vec{c} = 4 \vec{b}\).

### Step 2: Rearranging the equation
From the equation \(\vec{a} + \lambda \vec{c} = 4 \vec{b}\), we can express \(\vec{a}\) in terms of \(\vec{b}\) and \(\vec{c}\):
\[ \vec{a} = 4 \vec{b} - \lambda \vec{c} \]

### Step 3: Taking the modulus
Next, we take the modulus of both sides:
\[ |\vec{a}| = |4 \vec{b} - \lambda \vec{c}| \]
Given that \(|\vec{a}| = 3\), we have:
\[ 3 = |4 \vec{b} - \lambda \vec{c}| \]

### Step 4: Squaring both sides
To eliminate the modulus, we square both sides:
\[ 9 = |4 \vec{b} - \lambda \vec{c}|^2 \]

### Step 5: Expanding the squared modulus
Using the property of the modulus of a vector, we expand the right-hand side:
\[ |4 \vec{b} - \lambda \vec{c}|^2 = |4 \vec{b}|^2 + |\lambda \vec{c}|^2 - 2(4 \vec{b} \cdot \lambda \vec{c}) \]
Since \(|\vec{b}| = 1\) and \(|\vec{c}| = 1\) (because they are unit vectors), we have:
\[ |4 \vec{b}|^2 = 16, \quad |\lambda \vec{c}|^2 = \lambda^2 \]
Thus, the equation becomes:
\[ 9 = 16 + \lambda^2 - 8 \lambda (\vec{b} \cdot \vec{c}) \]

### Step 6: Substitute \(\vec{b} \cdot \vec{c}\)
Since \(\vec{b}\) and \(\vec{c}\) are parallel unit vectors, \(\vec{b} \cdot \vec{c} = 1\). Therefore, we substitute this into the equation:
\[ 9 = 16 + \lambda^2 - 8 \lambda \]

### Step 7: Rearranging the equation
Rearranging gives us:
\[ \lambda^2 - 8\lambda + 16 - 9 = 0 \]
This simplifies to:
\[ \lambda^2 - 8\lambda + 7 = 0 \]

### Step 8: Factoring the quadratic equation
We can factor the quadratic equation:
\[ (\lambda - 7)(\lambda - 1) = 0 \]
Thus, the solutions for \(\lambda\) are:
\[ \lambda = 7 \quad \text{and} \quad \lambda = 1 \]

### Step 9: Finding the sum of positive values
The problem asks for the sum of all possible positive values of \(\lambda\):
\[ \text{Sum} = 7 + 1 = 8 \]

### Final Answer
The sum of all possible positive values of \(\lambda\) is:
\[ \boxed{8} \]