What is the area of the triangle whose vertices are (0,0,0),(1,2,3) and (-3,-2,1) ?

Options

A. `3sqrt5` square unit
B. `6sqrt5` square unit
C. 6 square unit
D. 12 square unit

Explanation

To find the area of the triangle with vertices at points \( A(0,0,0) \), \( B(1,2,3) \), and \( C(-3,-2,1) \), we can use the formula for the area of a triangle formed by three points in 3D space.

### Step-by-Step Solution:

1. **Represent the Points as Vectors:**
- Let \( A = (0,0,0) \)
- Let \( B = (1,2,3) \)
- Let \( C = (-3,-2,1) \)

2. **Find Vectors AB and AC:**
- Vector \( \overrightarrow{AB} = B - A = (1 - 0, 2 - 0, 3 - 0) = (1, 2, 3) \)
- Vector \( \overrightarrow{AC} = C - A = (-3 - 0, -2 - 0, 1 - 0) = (-3, -2, 1) \)

3. **Calculate the Cross Product of Vectors AB and AC:**
- The cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \) can be calculated using the determinant of a matrix:
\[ \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ -3 & -2 & 1 \end{vmatrix} \]

4. **Calculate the Determinant:**
- Expanding the determinant:
\[ \overrightarrow{AB} \times \overrightarrow{AC} = \mathbf{i} \begin{vmatrix} 2 & 3 \\ -2 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 3 \\ -3 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 2 \\ -3 & -2 \end{vmatrix} \]
- Calculating the minors:
- For \( \mathbf{i} \): \( (2 \cdot 1 - 3 \cdot -2) = 2 + 6 = 8 \)
- For \( \mathbf{j} \): \( (1 \cdot 1 - 3 \cdot -3) = 1 + 9 = 10 \)
- For \( \mathbf{k} \): \( (1 \cdot -2 - 2 \cdot -3) = -2 + 6 = 4 \)
- Therefore:
\[ \overrightarrow{AB} \times \overrightarrow{AC} = 8\mathbf{i} - 10\mathbf{j} + 4\mathbf{k} \]

5. **Calculate the Magnitude of the Cross Product:**
- The magnitude is given by:
\[ |\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{8^2 + (-10)^2 + 4^2} = \sqrt{64 + 100 + 16} = \sqrt{180} = 6\sqrt{5} \]

6. **Calculate the Area of the Triangle:**
- The area \( A \) of the triangle is given by:
\[ A = \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{1}{2} \times 6\sqrt{5} = 3\sqrt{5} \]

### Final Answer:
The area of the triangle is \( 3\sqrt{5} \).