Find the area of the traingle ABC whose vertices are :
`A(1,2,4);B(-2,1,2)` and `C(2,4,-3)`.

Answer

To find the area of triangle ABC with vertices A(1, 2, 4), B(-2, 1, 2), and C(2, 4, -3), we can follow these steps:

### Step 1: Find the vectors AB and AC
We first need to find the vectors AB and AC.

- **Vector AB** is calculated as:
\[ \overrightarrow{AB} = B - A = (-2 - 1, 1 - 2, 2 - 4) = (-3, -1, -2) \]

- **Vector AC** is calculated as:
\[ \overrightarrow{AC} = C - A = (2 - 1, 4 - 2, -3 - 4) = (1, 2, -7) \]

### Step 2: Calculate the cross product of vectors AB and AC
Next, we calculate the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\).

Using the determinant formula for the cross product:
\[ \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & -1 & -2 \\ 1 & 2 & -7 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \left((-1)(-7) - (-2)(2)\right) - \hat{j} \left((-3)(-7) - (-2)(1)\right) + \hat{k} \left((-3)(2) - (-1)(1)\right) \]
\[ = \hat{i} (7 - (-4)) - \hat{j} (21 - (-2)) + \hat{k} (-6 - (-1)) \]
\[ = \hat{i} (7 + 4) - \hat{j} (21 + 2) + \hat{k} (-6 + 1) \]
\[ = 11\hat{i} - 23\hat{j} - 5\hat{k} \]

### Step 3: Find the magnitude of the cross product
Now we find the magnitude of the vector \(\overrightarrow{AB} \times \overrightarrow{AC}\):
\[ |\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(11)^2 + (-23)^2 + (-5)^2} \]
\[ = \sqrt{121 + 529 + 25} = \sqrt{675} = 15\sqrt{3} \]

### Step 4: Calculate the area of triangle ABC
The area of triangle ABC is given by:
\[ \text{Area} = \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}| \]
\[ = \frac{1}{2} (15\sqrt{3}) = \frac{15\sqrt{3}}{2} \]

### Final Answer
Thus, the area of triangle ABC is:
\[ \frac{15\sqrt{3}}{2} \text{ square units} \]

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