If `vec(A)=hat(i)+2hat(j)+3 hat(k), vec(B)=-hat(i)+hat(j)+4hat(k)` and `vec(C)=3hat(i)-3hat(j)-12 hat(k)`, then find the angle between the vectors `(vec(A)+vec(B)+vec(C))` and `(vec(A)xxvec(B))` in degrees.

Explanation

To find the angle between the vectors \( \vec{A} + \vec{B} + \vec{C} \) and \( \vec{A} \times \vec{B} \), we will follow these steps:

### Step 1: Calculate \( \vec{A} + \vec{B} + \vec{C} \)

Given:
\[ \vec{A} = \hat{i} + 2\hat{j} + 3\hat{k} \]
\[ \vec{B} = -\hat{i} + \hat{j} + 4\hat{k} \]
\[ \vec{C} = 3\hat{i} - 3\hat{j} - 12\hat{k} \]

Now, we will add these vectors together:
\[ \vec{A} + \vec{B} + \vec{C} = (\hat{i} - \hat{i} + 3\hat{i}) + (2\hat{j} + \hat{j} - 3\hat{j}) + (3\hat{k} + 4\hat{k} - 12\hat{k}) \]

Calculating each component:
- For \( \hat{i} \): \( 1 - 1 + 3 = 3 \)
- For \( \hat{j} \): \( 2 + 1 - 3 = 0 \)
- For \( \hat{k} \): \( 3 + 4 - 12 = -5 \)

Thus,
\[ \vec{A} + \vec{B} + \vec{C} = 3\hat{i} + 0\hat{j} - 5\hat{k} = 3\hat{i} - 5\hat{k} \]

### Step 2: Calculate \( \vec{A} \times \vec{B} \)

Using the determinant to find the cross product:
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -1 & 1 & 4 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} - \hat{j} \begin{vmatrix} 1 & 3 \\ -1 & 4 \end{vmatrix} + \hat{k} \begin{vmatrix} 1 & 2 \\ -1 & 1 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \( \begin{vmatrix} 2 & 3 \\ 1 & 4 \end{vmatrix} = (2)(4) - (3)(1) = 8 - 3 = 5 \)
2. \( \begin{vmatrix} 1 & 3 \\ -1 & 4 \end{vmatrix} = (1)(4) - (3)(-1) = 4 + 3 = 7 \)
3. \( \begin{vmatrix} 1 & 2 \\ -1 & 1 \end{vmatrix} = (1)(1) - (2)(-1) = 1 + 2 = 3 \)

Putting it all together:
\[ \vec{A} \times \vec{B} = 5\hat{i} - 7\hat{j} + 3\hat{k} \]

### Step 3: Calculate the angle between \( \vec{A} + \vec{B} + \vec{C} \) and \( \vec{A} \times \vec{B} \)

Let \( \vec{X} = \vec{A} + \vec{B} + \vec{C} = 3\hat{i} - 5\hat{k} \) and \( \vec{Y} = \vec{A} \times \vec{B} = 5\hat{i} - 7\hat{j} + 3\hat{k} \).

Using the formula for the angle \( \theta \) between two vectors:
\[ \cos \theta = \frac{\vec{X} \cdot \vec{Y}}{|\vec{X}| |\vec{Y}|} \]

### Step 4: Calculate \( \vec{X} \cdot \vec{Y} \)

Calculating the dot product:
\[ \vec{X} \cdot \vec{Y} = (3\hat{i} - 5\hat{k}) \cdot (5\hat{i} - 7\hat{j} + 3\hat{k}) = (3)(5) + (0)(-7) + (-5)(3) = 15 + 0 - 15 = 0 \]

### Step 5: Calculate the magnitudes \( |\vec{X}| \) and \( |\vec{Y}| \)

Calculating the magnitude of \( \vec{X} \):
\[ |\vec{X}| = \sqrt{3^2 + 0^2 + (-5)^2} = \sqrt{9 + 0 + 25} = \sqrt{34} \]

Calculating the magnitude of \( \vec{Y} \):
\[ |\vec{Y}| = \sqrt{5^2 + (-7)^2 + 3^2} = \sqrt{25 + 49 + 9} = \sqrt{83} \]

### Step 6: Find \( \cos \theta \)

Substituting into the cosine formula:
\[ \cos \theta = \frac{0}{\sqrt{34} \cdot \sqrt{83}} = 0 \]

### Step 7: Determine the angle \( \theta \)

Since \( \cos \theta = 0 \), this implies:
\[ \theta = 90^\circ \]

### Final Answer
The angle between the vectors \( \vec{A} + \vec{B} + \vec{C} \) and \( \vec{A} \times \vec{B} \) is \( 90^\circ \).