If `veca and vecb` are non zero, non collinear vectors and `veca_(1)=lambda veca+3 vecb, vecb_(1)=2veca+lambda vecb, vec c_(1)=veca+vecb`. Find the sum of all possible real values of `lambda` so that points `A_(1), B_(1), C_(1)` whose position vectors are `veca_(1), vecb_(1), vec c_(1)` respectively are collinear is equal to.

Answer

To solve the problem, we need to determine the values of \( \lambda \) such that the points \( A_1, B_1, C_1 \) with position vectors \( \vec{a_1}, \vec{b_1}, \vec{c_1} \) are collinear. The position vectors are given as follows:

\[ \vec{a_1} = \lambda \vec{a} + 3 \vec{b} \]
\[ \vec{b_1} = 2 \vec{a} + \lambda \vec{b} \]
\[ \vec{c_1} = \vec{a} + \vec{b} \]

### Step 1: Find the vectors \( \vec{A_1B_1} \) and \( \vec{A_1C_1} \)

The vector \( \vec{A_1B_1} \) can be expressed as:

\[ \vec{A_1B_1} = \vec{b_1} - \vec{a_1} = (2 \vec{a} + \lambda \vec{b}) - (\lambda \vec{a} + 3 \vec{b}) \]
\[ = (2 - \lambda) \vec{a} + (\lambda - 3) \vec{b} \]

The vector \( \vec{A_1C_1} \) can be expressed as:

\[ \vec{A_1C_1} = \vec{c_1} - \vec{a_1} = (\vec{a} + \vec{b}) - (\lambda \vec{a} + 3 \vec{b}) \]
\[ = (1 - \lambda) \vec{a} + (1 - 3) \vec{b} = (1 - \lambda) \vec{a} - 2 \vec{b} \]

### Step 2: Set up the condition for collinearity

The points \( A_1, B_1, C_1 \) are collinear if the vectors \( \vec{A_1B_1} \) and \( \vec{A_1C_1} \) are linearly dependent. This can be expressed using the cross product:

\[ \vec{A_1B_1} \times \vec{A_1C_1} = \vec{0} \]

Substituting the vectors we found:

\[ ((2 - \lambda) \vec{a} + (\lambda - 3) \vec{b}) \times ((1 - \lambda) \vec{a} - 2 \vec{b}) = \vec{0} \]

### Step 3: Calculate the cross product

Using the distributive property of the cross product, we have:

\[ = (2 - \lambda)(1 - \lambda) (\vec{a} \times \vec{a}) - 2(2 - \lambda)(\vec{a} \times \vec{b}) + (\lambda - 3)(1 - \lambda)(\vec{b} \times \vec{a}) - 2(\lambda - 3)(\vec{b} \times \vec{b}) \]

Since \( \vec{a} \times \vec{a} = \vec{0} \) and \( \vec{b} \times \vec{b} = \vec{0} \), we simplify to:

\[ -2(2 - \lambda)(\vec{a} \times \vec{b}) + (\lambda - 3)(1 - \lambda)(\vec{b} \times \vec{a}) = \vec{0} \]

Using the property \( \vec{b} \times \vec{a} = -(\vec{a} \times \vec{b}) \):

\[ -2(2 - \lambda)(\vec{a} \times \vec{b}) - (\lambda - 3)(1 - \lambda)(\vec{a} \times \vec{b}) = \vec{0} \]

Factoring out \( \vec{a} \times \vec{b} \):

\[ \left[-2(2 - \lambda) - (\lambda - 3)(1 - \lambda)\right] (\vec{a} \times \vec{b}) = \vec{0} \]

### Step 4: Solve the equation

Setting the coefficient to zero:

\[ -2(2 - \lambda) - (\lambda - 3)(1 - \lambda) = 0 \]

Expanding:

\[ -4 + 2\lambda - (\lambda - 3)(1 - \lambda) = 0 \]
\[ -4 + 2\lambda - (\lambda - 3 + \lambda^2 - 3\lambda) = 0 \]
\[ -4 + 2\lambda - \lambda^2 + \lambda + 3 = 0 \]
\[ -\lambda^2 + 3\lambda - 1 = 0 \]

Multiplying through by -1:

\[ \lambda^2 - 3\lambda + 1 = 0 \]

### Step 5: Find the roots

Using the quadratic formula:

\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} \]

### Step 6: Sum of all possible values of \( \lambda \)

The sum of the roots \( \lambda_1 + \lambda_2 \) is given by:

\[ \lambda_1 + \lambda_2 = 3 \]

Thus, the final answer is:

\[ \text{Sum of all possible real values of } \lambda = 3 \]