If `veca, vecb and vecc` are three non - zero and non - coplanar vectors such that `[(veca,vecb,vecc)]=4`, then the value of `(veca+3vecb-vecc).((veca-vecb)xx(veca-2vecb-3vecc))` equal to

Options

A. 40
B. 44
C. 48
D. 52

Answer

To solve the problem, we need to evaluate the expression \((\vec{a} + 3\vec{b} - \vec{c}) \cdot ((\vec{a} - \vec{b}) \times (\vec{a} - 2\vec{b} - 3\vec{c}))\) given that \([\vec{a}, \vec{b}, \vec{c}] = 4\).

### Step 1: Identify the vectors
We have three non-zero and non-coplanar vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). The scalar triple product \([\vec{a}, \vec{b}, \vec{c}] = 4\) indicates the volume of the parallelepiped formed by these vectors.

### Step 2: Rewrite the expression
We need to evaluate:
\[ (\vec{a} + 3\vec{b} - \vec{c}) \cdot ((\vec{a} - \vec{b}) \times (\vec{a} - 2\vec{b} - 3\vec{c})) \]

### Step 3: Calculate the cross product
To find \((\vec{a} - \vec{b}) \times (\vec{a} - 2\vec{b} - 3\vec{c})\), we can use the distributive property of the cross product:
\[ \vec{u} \times (\vec{v} + \vec{w}) = \vec{u} \times \vec{v} + \vec{u} \times \vec{w} \]
Let \(\vec{u} = \vec{a} - \vec{b}\), \(\vec{v} = \vec{a} - 2\vec{b}\), and \(\vec{w} = -3\vec{c}\).

Thus,
\[ (\vec{a} - \vec{b}) \times (\vec{a} - 2\vec{b} - 3\vec{c}) = (\vec{a} - \vec{b}) \times (\vec{a} - 2\vec{b}) + (\vec{a} - \vec{b}) \times (-3\vec{c}) \]

### Step 4: Simplify the expression
We can simplify the expression further by calculating each component:
1. \((\vec{a} - \vec{b}) \times (\vec{a} - 2\vec{b})\)
2. \((\vec{a} - \vec{b}) \times (-3\vec{c}) = -3(\vec{a} - \vec{b}) \times \vec{c}\)

### Step 5: Use the scalar triple product
The scalar triple product can be used to simplify the calculations. We know that:
\[ \vec{u} \cdot (\vec{v} \times \vec{w}) = [\vec{u}, \vec{v}, \vec{w}] \]
Thus, we can express our dot product as:
\[ (\vec{a} + 3\vec{b} - \vec{c}) \cdot ((\vec{a} - \vec{b}) \times (\vec{a} - 2\vec{b} - 3\vec{c})) = [\vec{a} + 3\vec{b} - \vec{c}, \vec{a} - \vec{b}, \vec{a} - 2\vec{b} - 3\vec{c}] \]

### Step 6: Calculate the determinant
The determinant can be calculated using the coefficients of the vectors:
\[ \begin{vmatrix} 1 & 3 & -1 \\ 1 & -1 & 0 \\ 1 & -2 & -3 \end{vmatrix} \]

### Step 7: Expand the determinant
Expanding the determinant:
1. The first row gives:
\[ 1 \cdot \begin{vmatrix} -1 & 0 \\ -2 & -3 \end{vmatrix} - 3 \cdot \begin{vmatrix} 1 & 0 \\ 1 & -3 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 1 & -1 \\ 1 & -2 \end{vmatrix} \]

Calculating these determinants:
- \(\begin{vmatrix} -1 & 0 \\ -2 & -3 \end{vmatrix} = 3\)
- \(\begin{vmatrix} 1 & 0 \\ 1 & -3 \end{vmatrix} = -3\)
- \(\begin{vmatrix} 1 & -1 \\ 1 & -2 \end{vmatrix} = -1\)

Putting it all together:
\[ 1 \cdot 3 - 3 \cdot (-3) + 1 \cdot (-1) = 3 + 9 - 1 = 11 \]

### Step 8: Multiply by the scalar triple product
Finally, we multiply by the scalar triple product:
\[ 11 \times 4 = 44 \]

Thus, the value of the expression is:
\[ \boxed{44} \]