To solve the problem, we need to calculate the scalar triple product of the vectors \(\vec{a} + \vec{b} + \vec{c}\), \(\vec{b} - \vec{a}\), and \(\vec{c}\). Given that \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are mutually perpendicular vectors with magnitudes 1, 2, and 3 respectively, we can proceed as follows:
### Step 1: Understand the Vectors
We know:
- \(|\vec{a}| = 1\)
- \(|\vec{b}| = 2\)
- \(|\vec{c}| = 3\)
Since the vectors are mutually perpendicular, we can represent them in a Cartesian coordinate system:
- Let \(\vec{a} = (1, 0, 0)\)
- Let \(\vec{b} = (0, 2, 0)\)
- Let \(\vec{c} = (0, 0, 3)\)
### Step 2: Calculate \(\vec{a} + \vec{b} + \vec{c}\)
\[
\vec{a} + \vec{b} + \vec{c} = (1, 0, 0) + (0, 2, 0) + (0, 0, 3) = (1, 2, 3)
\]
### Step 3: Calculate \(\vec{b} - \vec{a}\)
\[
\vec{b} - \vec{a} = (0, 2, 0) - (1, 0, 0) = (-1, 2, 0)
\]
### Step 4: Set Up the Scalar Triple Product
The scalar triple product can be computed as:
\[
[\vec{u}, \vec{v}, \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})
\]
where \(\vec{u} = \vec{a} + \vec{b} + \vec{c}\), \(\vec{v} = \vec{b} - \vec{a}\), and \(\vec{w} = \vec{c}\).
### Step 5: Calculate \(\vec{v} \times \vec{w}\)
First, we need to compute \(\vec{v} \times \vec{w}\):
\[
\vec{v} = (-1, 2, 0), \quad \vec{w} = (0, 0, 3)
\]
Using the determinant to find the cross product:
\[
\vec{v} \times \vec{w} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
-1 & 2 & 0 \\
0 & 0 & 3
\end{vmatrix}
= \hat{i}(2 \cdot 3 - 0 \cdot 0) - \hat{j}(-1 \cdot 3 - 0 \cdot 0) + \hat{k}(-1 \cdot 0 - 2 \cdot 0)
\]
\[
= 6\hat{i} + 3\hat{j} + 0\hat{k} = (6, 3, 0)
\]
### Step 6: Calculate \(\vec{u} \cdot (\vec{v} \times \vec{w})\)
Now, we compute the dot product:
\[
\vec{u} = (1, 2, 3)
\]
\[
\vec{v} \times \vec{w} = (6, 3, 0)
\]
\[
\vec{u} \cdot (\vec{v} \times \vec{w}) = (1, 2, 3) \cdot (6, 3, 0) = 1 \cdot 6 + 2 \cdot 3 + 3 \cdot 0 = 6 + 6 + 0 = 12
\]
### Final Answer
Thus, the value of the scalar triple product is:
\[
[\vec{a} + \vec{b} + \vec{c}, \vec{b} - \vec{a}, \vec{c}] = 12
\]
