p=2a-3b,q=a-2b+c and r=-3a+b+2c, where a,b,c being non-coplanar vectors, then the vector -2a+3b-c is equal to

Options

A. `p-4q`
B. `(-7q+r)/(5)`
C. `2p-3q+r`
D. `4p-2r`

Explanation

To solve the problem, we need to express the vector \(-2a + 3b - c\) in terms of the vectors \(p\), \(q\), and \(r\).

We start with the definitions of the vectors:
- \(p = 2a + 3b\)
- \(q = a - 2b + c\)
- \(r = -3a + b + 2c\)

We want to express \(-2a + 3b - c\) in the form \(xp + yq + zr\).

### Step 1: Set up the equation
We can write:
\[ -2a + 3b - c = xp + yq + zr \]
Substituting the expressions for \(p\), \(q\), and \(r\):
\[ -2a + 3b - c = x(2a + 3b) + y(a - 2b + c) + z(-3a + b + 2c) \]

### Step 2: Expand the right-hand side
Expanding the right side gives:
\[ -2a + 3b - c = (2x + y - 3z)a + (3x - 2y + z)b + (y + 2z)c \]

### Step 3: Equate coefficients
Now, we can equate coefficients of \(a\), \(b\), and \(c\) from both sides:
1. For \(a\): \(2x + y - 3z = -2\) (Equation 1)
2. For \(b\): \(3x - 2y + z = 3\) (Equation 2)
3. For \(c\): \(y + 2z = -1\) (Equation 3)

### Step 4: Solve the system of equations
We will solve these three equations step by step.

From Equation 3:
\[ y + 2z = -1 \implies y = -1 - 2z \quad (Equation 4) \]

Substituting Equation 4 into Equation 1:
\[ 2x + (-1 - 2z) - 3z = -2 \]
\[ 2x - 1 - 5z = -2 \implies 2x - 5z = -1 \implies 2x = 5z - 1 \implies x = \frac{5z - 1}{2} \quad (Equation 5) \]

Now substitute Equation 4 into Equation 2:
\[ 3x - 2(-1 - 2z) + z = 3 \]
\[ 3x + 2 + 4z + z = 3 \implies 3x + 5z = 1 \implies 3x = 1 - 5z \implies x = \frac{1 - 5z}{3} \quad (Equation 6) \]

### Step 5: Equate Equations 5 and 6
Now we have two expressions for \(x\):
\[ \frac{5z - 1}{2} = \frac{1 - 5z}{3} \]

Cross-multiplying gives:
\[ 3(5z - 1) = 2(1 - 5z) \]
\[ 15z - 3 = 2 - 10z \]
\[ 15z + 10z = 2 + 3 \implies 25z = 5 \implies z = \frac{1}{5} \]

### Step 6: Find \(y\) and \(x\)
Substituting \(z = \frac{1}{5}\) into Equation 4:
\[ y = -1 - 2\left(\frac{1}{5}\right) = -1 - \frac{2}{5} = -\frac{5}{5} - \frac{2}{5} = -\frac{7}{5} \]

Now substituting \(z = \frac{1}{5}\) into Equation 5:
\[ x = \frac{5\left(\frac{1}{5}\right) - 1}{2} = \frac{1 - 1}{2} = 0 \]

### Step 7: Final expression
Now we have:
- \(x = 0\)
- \(y = -\frac{7}{5}\)
- \(z = \frac{1}{5}\)

Thus, we can express \(-2a + 3b - c\) as:
\[ -2a + 3b - c = 0 \cdot p - \frac{7}{5}q + \frac{1}{5}r \]
This simplifies to:
\[ -2a + 3b - c = -\frac{7}{5}q + \frac{1}{5}r \]

### Conclusion
The vector \(-2a + 3b - c\) is equal to:
\[ -\frac{7}{5}q + \frac{1}{5}r \]