The vector equation of the plane passing through ` vec a ,\ vec b , vec c` is ` vec r=alpha vec a+beta vec b+gamma vec c` provided that

Options

A. a) `alpha+beta+gamma=0`
B. b) `alpha+beta+gamma=1`
C. c) `alpha+beta=gamma`
D. d) `alpha^2+beta^2+gamma^2=1`

Explanation

To solve the problem of finding the value of \( \alpha + \beta + \gamma \) in the vector equation of the plane given by \( \vec{r} = \alpha \vec{a} + \beta \vec{b} + \gamma \vec{c} \), we can follow these steps:

### Step 1: Understand the Vector Equation of the Plane
The vector equation of a plane passing through three points \( \vec{a}, \vec{b}, \vec{c} \) can be expressed in terms of parameters \( \alpha, \beta, \) and \( \gamma \). The equation is given as:
\[ \vec{r} = \alpha \vec{a} + \beta \vec{b} + \gamma \vec{c} \]

### Step 2: Use the Properties of the Plane
Since \( \vec{a}, \vec{b}, \vec{c} \) are points on the plane, any point \( \vec{r} \) on the plane can be represented as a linear combination of these points. The coefficients \( \alpha, \beta, \gamma \) represent the weights of the respective position vectors.

### Step 3: Apply the Condition for the Plane
For any point \( \vec{r} \) on the plane formed by \( \vec{a}, \vec{b}, \vec{c} \), the sum of the coefficients must equal 1. This is because the coefficients represent a weighted average of the position vectors, and they must sum to 1 to ensure that \( \vec{r} \) lies within the triangle formed by \( \vec{a}, \vec{b}, \vec{c} \).

Thus, we have the condition:
\[ \alpha + \beta + \gamma = 1 \]

### Conclusion
Therefore, the value of \( \alpha + \beta + \gamma \) is:
\[ \alpha + \beta + \gamma = 1 \]

### Final Answer
The answer is \( \alpha + \beta + \gamma = 1 \).

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