Let ` vec a , vec b ,vec c ` are three non- coplanar vectors such that
`vecr_(1)=veca + vec c , vecr_(2)= vec b+vec c -veca , vec r_(3) = vec c + vec a + vecb, vec r = 2 vec a - 3 vec b + 4 vec c. `
If `vec r = lambda_(1)vecr_(1)+lambda_(2)vecr_(2)+lambda_(3)vecr_(3)`, then

Options

A. `lambda_(1)=7`
B. `lambda_(1)+lambda_(3)=3`
C. `lambda_(1)+lambda_(2)+lambda_(3)=3`
D. `lambda_(3)+lambda_(2) =2`

Answer

To solve the given problem, we need to express the vector \( \vec{r} \) in terms of the vectors \( \vec{r}_1, \vec{r}_2, \) and \( \vec{r}_3 \) and find the values of \( \lambda_1, \lambda_2, \) and \( \lambda_3 \).

### Step 1: Write down the given vectors
We have the following vectors defined:
- \( \vec{r}_1 = \vec{a} + \vec{c} \)
- \( \vec{r}_2 = \vec{b} + \vec{c} - \vec{a} \)
- \( \vec{r}_3 = \vec{c} + \vec{a} + \vec{b} \)
- \( \vec{r} = 2\vec{a} - 3\vec{b} + 4\vec{c} \)

### Step 2: Set up the equation
We need to express \( \vec{r} \) as a linear combination of \( \vec{r}_1, \vec{r}_2, \) and \( \vec{r}_3 \):
\[ \vec{r} = \lambda_1 \vec{r}_1 + \lambda_2 \vec{r}_2 + \lambda_3 \vec{r}_3 \]

### Step 3: Substitute the expressions for \( \vec{r}_1, \vec{r}_2, \) and \( \vec{r}_3 \)
Substituting the expressions for \( \vec{r}_1, \vec{r}_2, \) and \( \vec{r}_3 \) into the equation, we get:
\[ 2\vec{a} - 3\vec{b} + 4\vec{c} = \lambda_1 (\vec{a} + \vec{c}) + \lambda_2 (\vec{b} + \vec{c} - \vec{a}) + \lambda_3 (\vec{c} + \vec{a} + \vec{b}) \]

### Step 4: Expand the right-hand side
Expanding the right-hand side:
\[ = \lambda_1 \vec{a} + \lambda_1 \vec{c} + \lambda_2 \vec{b} + \lambda_2 \vec{c} - \lambda_2 \vec{a} + \lambda_3 \vec{c} + \lambda_3 \vec{a} + \lambda_3 \vec{b} \]
Combining like terms:
\[ = (\lambda_1 - \lambda_2 + \lambda_3) \vec{a} + (\lambda_2 + \lambda_3) \vec{b} + (\lambda_1 + \lambda_2 + \lambda_3) \vec{c} \]

### Step 5: Equate coefficients
Now we equate the coefficients of \( \vec{a}, \vec{b}, \) and \( \vec{c} \) from both sides:
1. For \( \vec{a} \):
\[ \lambda_1 - \lambda_2 + \lambda_3 = 2 \quad (1) \]
2. For \( \vec{b} \):
\[ \lambda_2 + \lambda_3 = -3 \quad (2) \]
3. For \( \vec{c} \):
\[ \lambda_1 + \lambda_2 + \lambda_3 = 4 \quad (3) \]

### Step 6: Solve the system of equations
From equation (2):
\[ \lambda_3 = -3 - \lambda_2 \quad (4) \]
Substituting (4) into (1):
\[ \lambda_1 - \lambda_2 + (-3 - \lambda_2) = 2 \]
\[ \lambda_1 - 2\lambda_2 - 3 = 2 \]
\[ \lambda_1 - 2\lambda_2 = 5 \quad (5) \]

Now substituting (4) into (3):
\[ \lambda_1 + \lambda_2 + (-3 - \lambda_2) = 4 \]
\[ \lambda_1 - 3 = 4 \]
\[ \lambda_1 = 7 \quad (6) \]

### Step 7: Substitute \( \lambda_1 \) back to find \( \lambda_2 \) and \( \lambda_3 \)
Substituting (6) into (5):
\[ 7 - 2\lambda_2 = 5 \]
\[ -2\lambda_2 = -2 \]
\[ \lambda_2 = 1 \quad (7) \]

Now substituting (7) into (4):
\[ \lambda_3 = -3 - 1 = -4 \quad (8) \]

### Final Values
Thus, we have:
- \( \lambda_1 = 7 \)
- \( \lambda_2 = 1 \)
- \( \lambda_3 = -4 \)

### Conclusion
The values of \( \lambda_1, \lambda_2, \) and \( \lambda_3 \) are \( 7, 1, \) and \( -4 \) respectively.