Volume of parallelepiped determined by vectors `bara,barb` and `barc` is 5.
Then the volume of the parallelepiped determined by the vectors `3(bara +barb). (barb + barc)` and 2(`barc + bara)` is

Options

A. `3(bara+barb),(barb+barc)` and `(barc+bara)` is 24
B. `2(baraxxbarb),3(barbxxbarc)` and `(barcxxbara)` is 24
C. `(bara-2barb),(barb-2barc)` and `(barc-2bara)` is 14
D. `3(bara-barb),(3barb-barc)` and `(3barc-bara)` is 52

Answer

To find the volume of the parallelepiped determined by the vectors \(3(\mathbf{a} + \mathbf{b})\), \((\mathbf{b} + \mathbf{c})\), and \(2(\mathbf{c} + \mathbf{a})\), we can use the properties of the scalar triple product (also known as the box product).

### Step-by-step Solution:

1. **Understand the Volume of Parallelepiped**:
The volume \(V\) of a parallelepiped formed by vectors \(\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{w}\) can be calculated using the scalar triple product:
\[ V = |\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})| \]

2. **Identify the Given Volume**:
We are given that the volume of the parallelepiped determined by vectors \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) is 5:
\[ V = |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| = 5 \]

3. **Express the New Vectors**:
We need to find the volume of the parallelepiped formed by the vectors:
\[ \mathbf{u} = 3(\mathbf{a} + \mathbf{b}), \quad \mathbf{v} = \mathbf{b} + \mathbf{c}, \quad \mathbf{w} = 2(\mathbf{c} + \mathbf{a}) \]

4. **Calculate the Scalar Triple Product**:
The volume can be expressed as:
\[ V' = |\mathbf{u} \cdot (\mathbf{v} \times \mathbf{w})| \]
Substituting the expressions for \(\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{w}\):
\[ V' = |3(\mathbf{a} + \mathbf{b}) \cdot ((\mathbf{b} + \mathbf{c}) \times (2(\mathbf{c} + \mathbf{a}))| \]

5. **Simplify the Cross Product**:
First, calculate the cross product:
\[ (\mathbf{b} + \mathbf{c}) \times (2(\mathbf{c} + \mathbf{a})) = 2[(\mathbf{b} + \mathbf{c}) \times \mathbf{c} + (\mathbf{b} + \mathbf{c}) \times \mathbf{a}] \]
Since \(\mathbf{c} \times \mathbf{c} = \mathbf{0}\):
\[ = 2[(\mathbf{b} \times \mathbf{a}) + (\mathbf{c} \times \mathbf{a})] \]

6. **Substitute Back into the Volume Expression**:
Now substitute back into the volume expression:
\[ V' = |3(\mathbf{a} + \mathbf{b}) \cdot (2[(\mathbf{b} \times \mathbf{a}) + (\mathbf{c} \times \mathbf{a})])| \]
\[ = 6 |(\mathbf{a} + \mathbf{b}) \cdot [(\mathbf{b} \times \mathbf{a}) + (\mathbf{c} \times \mathbf{a})]| \]

7. **Expand the Dot Product**:
\[ = 6 \left[ (\mathbf{a} \cdot (\mathbf{b} \times \mathbf{a})) + (\mathbf{b} \cdot (\mathbf{b} \times \mathbf{a})) + (\mathbf{a} \cdot (\mathbf{c} \times \mathbf{a})) + (\mathbf{b} \cdot (\mathbf{c} \times \mathbf{a})) \right] \]
The terms \(\mathbf{a} \cdot (\mathbf{b} \times \mathbf{a})\) and \(\mathbf{b} \cdot (\mathbf{b} \times \mathbf{a})\) are zero because the dot product of a vector with itself or with a vector that is perpendicular to it is zero.

8. **Final Calculation**:
Thus, we are left with:
\[ = 6 |(\mathbf{b} \cdot (\mathbf{c} \times \mathbf{a}))| \]
Since \(|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})| = 5\), we have:
\[ = 6 \times 5 = 30 \]

### Final Answer:
The volume of the parallelepiped determined by the vectors \(3(\mathbf{a} + \mathbf{b})\), \((\mathbf{b} + \mathbf{c})\), and \(2(\mathbf{c} + \mathbf{a})\) is \(30\).