Let `veca,vecb,vecc` be three linearly independent vectors, then `([veca+2vecb-vecc 2veca+vecb+vecc4veca-vecb+5vecc])/([vecavecbvecc])`

Options

A. 0
B. 1
C. 2
D. `-1`

Explanation

To solve the problem, we need to evaluate the expression given by the scalar triple product of the vectors in the numerator divided by the scalar triple product of the vectors in the denominator. Let's break it down step-by-step.

### Step 1: Define the vectors
Let:
- \(\vec{x} = \vec{a} + 2\vec{b} - \vec{c}\)
- \(\vec{y} = 2\vec{a} + \vec{b} + \vec{c}\)
- \(\vec{z} = 4\vec{a} - \vec{b} + 5\vec{c}\)

### Step 2: Set up the scalar triple product
The scalar triple product \([\vec{x}, \vec{y}, \vec{z}]\) can be represented as the determinant of a matrix formed by these vectors:
\[ [\vec{x}, \vec{y}, \vec{z}] = \begin{vmatrix} 1 & 2 & -1 \\ 2 & 1 & 1 \\ 4 & -1 & 5 \end{vmatrix} \]

### Step 3: Calculate the determinant
Now we will compute the determinant:
\[ \begin{vmatrix} 1 & 2 & -1 \\ 2 & 1 & 1 \\ 4 & -1 & 5 \end{vmatrix} \]

Using the formula for the determinant of a 3x3 matrix:
\[ \text{det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where the matrix is:
\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} \]

In our case:
- \(a = 1, b = 2, c = -1\)
- \(d = 2, e = 1, f = 1\)
- \(g = 4, h = -1, i = 5\)

Calculating the determinant:
\[ = 1(1 \cdot 5 - 1 \cdot (-1)) - 2(2 \cdot 5 - 1 \cdot 4) + (-1)(2 \cdot (-1) - 1 \cdot 4) \]
\[ = 1(5 + 1) - 2(10 - 4) - (2 - 4) \]
\[ = 1 \cdot 6 - 2 \cdot 6 + 2 \]
\[ = 6 - 12 + 2 = -4 \]

### Step 4: Evaluate the denominator
The denominator is the scalar triple product \([\vec{a}, \vec{b}, \vec{c}]\), which is non-zero since \(\vec{a}, \vec{b}, \vec{c}\) are linearly independent vectors. Let's denote this value as \(D\).

### Step 5: Final expression
Now, we can express the original question as:
\[ \frac{[\vec{x}, \vec{y}, \vec{z}]}{[\vec{a}, \vec{b}, \vec{c}]} = \frac{-4}{D} \]

Since \(D\) is non-zero, the final answer is:
\[ \frac{-4}{D} \]

### Conclusion
The final result of the expression is \(\frac{-4}{D}\), where \(D\) is the scalar triple product of the vectors \(\vec{a}, \vec{b}, \vec{c}\).