`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. (a) `p-4q`
B. (b) `(-7q+r)/(5)`
C. (c) `2p-3q+r`
D. (d) `4p-2r`

Explanation

To solve the problem, we need to express the vector \(-2a + 3b - c\) as a linear combination of the vectors \(p\), \(q\), and \(r\). The vectors are defined as follows:

\[ p = 2a - 3b \]
\[ q = a - 2b + c \]
\[ r = -3a + b + 2c \]

We want to find constants \(l\), \(m\), and \(n\) such that:

\[ -2a + 3b - c = l \cdot p + m \cdot q + n \cdot r \]

### Step 1: Substitute \(p\), \(q\), and \(r\)

Substituting the expressions for \(p\), \(q\), and \(r\) into the equation gives:

\[ -2a + 3b - c = l(2a - 3b) + m(a - 2b + c) + n(-3a + b + 2c) \]

### Step 2: Expand the right-hand side

Expanding the right-hand side:

\[ = l(2a) - l(3b) + m(a) - m(2b) + m(c) + n(-3a) + n(b) + n(2c) \]

Combining like terms:

\[ = (2l + m - 3n)a + (-3l - 2m + n)b + (m + 2n)c \]

### Step 3: Set up equations for coefficients

Now we can equate the coefficients of \(a\), \(b\), and \(c\) from both sides:

1. Coefficient of \(a\):
\[ 2l + m - 3n = -2 \quad \text{(Equation 1)} \]

2. Coefficient of \(b\):
\[ -3l - 2m + n = 3 \quad \text{(Equation 2)} \]

3. Coefficient of \(c\):
\[ m + 2n = -1 \quad \text{(Equation 3)} \]

### Step 4: Solve the system of equations

From Equation 3, we can express \(m\) in terms of \(n\):

\[ m = -1 - 2n \]

Substituting \(m\) into Equations 1 and 2:

**Substituting into Equation 1:**
\[ 2l + (-1 - 2n) - 3n = -2 \]
\[ 2l - 1 - 5n = -2 \implies 2l - 5n = -1 \quad \text{(Equation 4)} \]

**Substituting into Equation 2:**
\[ -3l - 2(-1 - 2n) + n = 3 \]
\[ -3l + 2 + 4n + n = 3 \implies -3l + 5n = 1 \quad \text{(Equation 5)} \]

### Step 5: Solve Equations 4 and 5

Now we have a system of two equations:

1. \(2l - 5n = -1\) (Equation 4)
2. \(-3l + 5n = 1\) (Equation 5)

We can add these two equations to eliminate \(n\):

\[ (2l - 5n) + (-3l + 5n) = -1 + 1 \]
\[ -1l = 0 \implies l = 0 \]

Substituting \(l = 0\) back into Equation 4:

\[ 2(0) - 5n = -1 \implies -5n = -1 \implies n = \frac{1}{5} \]

Now substituting \(n = \frac{1}{5}\) back into the expression for \(m\):

\[ m = -1 - 2\left(\frac{1}{5}\right) = -1 - \frac{2}{5} = -\frac{7}{5} \]

### Step 6: Final Result

We have found:

\[ l = 0, \quad m = -\frac{7}{5}, \quad n = \frac{1}{5} \]

Thus, we can express \(-2a + 3b - c\) as:

\[ -2a + 3b - c = 0 \cdot p - \frac{7}{5} \cdot q + \frac{1}{5} \cdot r \]

This simplifies to:

\[ -2a + 3b - c = -\frac{7}{5}q + \frac{1}{5}r \]

### Conclusion

The vector \(-2a + 3b - c\) is equal to:

\[ -\frac{7}{5}q + \frac{1}{5}r \]