Let a, b, c be three non-coplanar vectors such that `r_(1) = a-b + c, r_(2) = b + c-a, r_(3) =c + a + b`
`r = 2a -3b + 4c`. If `r = p_(1)r_(1) + P_(2)r_(2) + P_(3)r_(3)`, then

Options

A. `p_(1) = 7`
B. `p_(1) + p_(3) = 3`
C. `p_(1) + p_(2) + p_(3) = 4`
D. `p_(3) + p_(2) = 0`

Answer

To solve the problem, we need to express the vector \( r \) in terms of the vectors \( r_1, r_2, \) and \( r_3 \) and find the coefficients \( P_1, P_2, \) and \( P_3 \).

### Step 1: Write down the given vectors
We have:
- \( r_1 = a - b + c \)
- \( r_2 = b + c - a \)
- \( r_3 = c + a + b \)
- \( r = 2a - 3b + 4c \)

### Step 2: Express \( r \) in terms of \( r_1, r_2, \) and \( r_3 \)
We need to express \( r \) as:
\[ r = P_1 r_1 + P_2 r_2 + P_3 r_3 \]
Substituting the values of \( r_1, r_2, \) and \( r_3 \):
\[ r = P_1 (a - b + c) + P_2 (b + c - a) + P_3 (c + a + b) \]

### Step 3: Expand the expression
Expanding the right-hand side:
\[ r = P_1 a - P_1 b + P_1 c + P_2 b + P_2 c - P_2 a + P_3 c + P_3 a + P_3 b \]
Combining like terms, we get:
\[ r = (P_1 - P_2 + P_3) a + (-P_1 + P_2 + P_3) b + (P_1 + P_2 + P_3) c \]

### Step 4: Set coefficients equal to those in \( r \)
Now we can compare coefficients with \( r = 2a - 3b + 4c \):
1. For \( a \): \( P_1 - P_2 + P_3 = 2 \) (Equation 1)
2. For \( b \): \( -P_1 + P_2 + P_3 = -3 \) (Equation 2)
3. For \( c \): \( P_1 + P_2 + P_3 = 4 \) (Equation 3)

### Step 5: Solve the system of equations
From Equation 3:
\[ P_1 + P_2 + P_3 = 4 \]
From Equation 1:
\[ P_1 - P_2 + P_3 = 2 \]
From Equation 2:
\[ -P_1 + P_2 + P_3 = -3 \]

### Step 6: Simplify the equations
We can subtract Equation 1 from Equation 3:
\[ (P_1 + P_2 + P_3) - (P_1 - P_2 + P_3) = 4 - 2 \]
This simplifies to:
\[ 2P_2 = 2 \implies P_2 = 1 \]

Now substituting \( P_2 = 1 \) into Equation 3:
\[ P_1 + 1 + P_3 = 4 \implies P_1 + P_3 = 3 \quad (Equation 4) \]

Substituting \( P_2 = 1 \) into Equation 2:
\[ -P_1 + 1 + P_3 = -3 \implies -P_1 + P_3 = -4 \implies P_3 = P_1 - 4 \quad (Equation 5) \]

### Step 7: Substitute Equation 5 into Equation 4
Substituting \( P_3 = P_1 - 4 \) into Equation 4:
\[ P_1 + (P_1 - 4) = 3 \implies 2P_1 - 4 = 3 \implies 2P_1 = 7 \implies P_1 = \frac{7}{2} \]

Now substituting \( P_1 = \frac{7}{2} \) back into Equation 5:
\[ P_3 = \frac{7}{2} - 4 = \frac{7}{2} - \frac{8}{2} = -\frac{1}{2} \]

### Step 8: Summary of the coefficients
Thus, we have:
- \( P_1 = \frac{7}{2} \)
- \( P_2 = 1 \)
- \( P_3 = -\frac{1}{2} \)

### Final Answer
The values of \( P_1, P_2, \) and \( P_3 \) are:
\[ P_1 = \frac{7}{2}, \quad P_2 = 1, \quad P_3 = -\frac{1}{2} \]