Given the following vectors
`r_(1) =( 2, -1, 1) " " r_(2) =( 1, 3,-2`)
`r_(3) =(-2, 1,-3) " "r_(4) =(3, 2,-5)`.
If `r_(4) = ar_(1) + br_(2) + cr_(3)`, then

Options

A. `a = b = c`
B. `b = (2ac)/(a + c)`
C. `a +b =2 c`
D. ` a= bc`

Answer

To solve the problem, we need to express the vector \( r_4 \) as a linear combination of the vectors \( r_1 \), \( r_2 \), and \( r_3 \). We will find the coefficients \( a \), \( b \), and \( c \) such that:

\[ r_4 = a r_1 + b r_2 + c r_3 \]

### Step 1: Write down the vectors
The given vectors are:
- \( r_1 = (2, -1, 1) \)
- \( r_2 = (1, 3, -2) \)
- \( r_3 = (-2, 1, -3) \)
- \( r_4 = (3, 2, -5) \)

### Step 2: Set up the equation
We can express the equation as follows:

\[ (3, 2, -5) = a(2, -1, 1) + b(1, 3, -2) + c(-2, 1, -3) \]

### Step 3: Expand the right-hand side
Expanding the right-hand side gives us:

\[ (3, 2, -5) = (2a + b - 2c, -a + 3b + c, a - 2b - 3c) \]

### Step 4: Set up the system of equations
From the above equation, we can create a system of equations by equating the corresponding components:

1. \( 2a + b - 2c = 3 \) (Equation 1)
2. \( -a + 3b + c = 2 \) (Equation 2)
3. \( a - 2b - 3c = -5 \) (Equation 3)

### Step 5: Solve the system of equations
We will solve these equations step by step.

#### Step 5.1: Solve Equation 1 for \( b \)
From Equation 1:
\[ b = 3 - 2a + 2c \]

#### Step 5.2: Substitute \( b \) in Equation 2
Substituting \( b \) into Equation 2:
\[ -a + 3(3 - 2a + 2c) + c = 2 \]
Expanding this:
\[ -a + 9 - 6a + 6c + c = 2 \]
Combining like terms:
\[ -7a + 7c + 9 = 2 \]
This simplifies to:
\[ -7a + 7c = -7 \quad \Rightarrow \quad a - c = 1 \quad \Rightarrow \quad a = c + 1 \quad (Equation 4) \]

#### Step 5.3: Substitute \( a \) in Equation 3
Now substituting \( a \) from Equation 4 into Equation 3:
\[ (c + 1) - 2b - 3c = -5 \]
Substituting \( b \) from Equation 1:
\[ (c + 1) - 2(3 - 2(c + 1) + 2c) - 3c = -5 \]
This leads to a complex equation, so we will simplify step by step.

#### Step 5.4: Solve for \( c \)
We can also directly sum the equations to eliminate \( c \):
Adding all three equations:
\[ (2a + b - 2c) + (-a + 3b + c) + (a - 2b - 3c) = 3 + 2 - 5 \]
This simplifies to:
\[ (2a - a + a) + (b + 3b - 2b) + (-2c + c - 3c) = 0 \]
Thus:
\[ 2a + 2b - 4c = 0 \quad \Rightarrow \quad a + b = 2c \quad (Equation 5) \]

### Step 6: Final equations
Now we have:
1. \( a - c = 1 \) (Equation 4)
2. \( a + b = 2c \) (Equation 5)

### Step 7: Solve the final equations
Substituting \( c = a - 1 \) into Equation 5:
\[ a + b = 2(a - 1) \]
This leads to:
\[ a + b = 2a - 2 \quad \Rightarrow \quad b = a - 2 \]

### Step 8: Conclusion
Substituting \( b = a - 2 \) and \( c = a - 1 \) gives us the relationships between \( a \), \( b \), and \( c \).

Thus, the final relation is:
\[ a + 2b = 2c \]