Let co-ordinates of a point 'p' with respectto the system non-coplanar vectors `veca, vecb and vecc` is (3, 2, 1). Then, co-ordinates of 'p'with respect to the system of vectors `vec a+vec b+vec c,vec a-vecb+vec c. vec a+vec b- vec c`

Options

A. `(3//2, 1//2 , 1 ) `
B. `(3//2, 1 , 1//2)`
C. `(1//2, 3//2,1)`
D. none of these

Explanation

To solve the problem, we need to find the coordinates of the point \( P \) with respect to the new system of vectors \( \vec{a} + \vec{b} + \vec{c} \), \( \vec{a} - \vec{b} + \vec{c} \), and \( \vec{a} + \vec{b} - \vec{c} \) given that the coordinates of \( P \) with respect to the original system of non-coplanar vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) are \( (3, 2, 1) \).

### Step-by-Step Solution:

1. **Express the coordinates in terms of the original vectors**:
The coordinates of point \( P \) with respect to the original vectors can be expressed as:
\[ P = 3\vec{a} + 2\vec{b} + 1\vec{c} \]
This is our Equation (1).

2. **Define the new system of vectors**:
Let:
\[ \vec{u} = \vec{a} + \vec{b} + \vec{c} \]
\[ \vec{v} = \vec{a} - \vec{b} + \vec{c} \]
\[ \vec{w} = \vec{a} + \vec{b} - \vec{c} \]

3. **Express \( P \) in terms of the new vectors**:
We want to express \( P \) as:
\[ P = \alpha \vec{u} + \beta \vec{v} + \gamma \vec{w} \]
Expanding this gives:
\[ P = \alpha(\vec{a} + \vec{b} + \vec{c}) + \beta(\vec{a} - \vec{b} + \vec{c}) + \gamma(\vec{a} + \vec{b} - \vec{c}) \]
Simplifying this, we get:
\[ P = (\alpha + \beta + \gamma)\vec{a} + (\alpha - \beta + \gamma)\vec{b} + (\alpha + \beta - \gamma)\vec{c} \]
This is our Equation (2).

4. **Set up the equations by comparing coefficients**:
From Equations (1) and (2), we can equate the coefficients of \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \):
- For \( \vec{a} \):
\[ \alpha + \beta + \gamma = 3 \quad \text{(Equation 3)} \]
- For \( \vec{b} \):
\[ \alpha - \beta + \gamma = 2 \quad \text{(Equation 4)} \]
- For \( \vec{c} \):
\[ \alpha + \beta - \gamma = 1 \quad \text{(Equation 5)} \]

5. **Solve the system of equations**:
Now we have a system of three equations:
1. \( \alpha + \beta + \gamma = 3 \)
2. \( \alpha - \beta + \gamma = 2 \)
3. \( \alpha + \beta - \gamma = 1 \)

**Subtract Equation 4 from Equation 3**:
\[ (\alpha + \beta + \gamma) - (\alpha - \beta + \gamma) = 3 - 2 \]
This simplifies to:
\[ 2\beta = 1 \implies \beta = \frac{1}{2} \]

**Substitute \( \beta \) back into Equations 3 and 5**:
From Equation 3:
\[ \alpha + \frac{1}{2} + \gamma = 3 \implies \alpha + \gamma = 3 - \frac{1}{2} = \frac{5}{2} \quad \text{(Equation 6)} \]

From Equation 5:
\[ \alpha + \frac{1}{2} - \gamma = 1 \implies \alpha - \gamma = 1 - \frac{1}{2} = \frac{1}{2} \quad \text{(Equation 7)} \]

**Now solve Equations 6 and 7**:
Adding Equations 6 and 7:
\[ (\alpha + \gamma) + (\alpha - \gamma) = \frac{5}{2} + \frac{1}{2} \]
This gives:
\[ 2\alpha = 3 \implies \alpha = \frac{3}{2} \]

Substitute \( \alpha \) back into Equation 6:
\[ \frac{3}{2} + \gamma = \frac{5}{2} \implies \gamma = 1 \]

6. **Final coordinates**:
We have found:
\[ \alpha = \frac{3}{2}, \quad \beta = \frac{1}{2}, \quad \gamma = 1 \]
Therefore, the coordinates of point \( P \) with respect to the new system of vectors are:
\[ \left( \frac{3}{2}, \frac{1}{2}, 1 \right) \]