Find the absolute value of parameter `t` for which the area of the triangle whose vertices the `A(-1,1,2); B(1,2,3)a n dC(t,1,1)` is minimum.

Explanation

To find the absolute value of parameter \( t \) for which the area of the triangle with vertices \( A(-1,1,2) \), \( B(1,2,3) \), and \( C(t,1,1) \) is minimized, we can follow these steps:

### Step 1: Calculate the vectors \( \overrightarrow{AB} \) and \( \overrightarrow{AC} \)

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

The vector \( \overrightarrow{AC} \) is calculated as:
\[ \overrightarrow{AC} = C - A = (t - (-1), 1 - 1, 1 - 2) = (t + 1, 0, -1) \]

### Step 2: Compute the cross product \( \overrightarrow{AB} \times \overrightarrow{AC} \)

Using the determinant to find the cross product:
\[ \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ t + 1 & 0 & -1 \end{vmatrix} \]

Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 1 & 1 \\ 0 & -1 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 1 \\ t + 1 & -1 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 1 \\ t + 1 & 0 \end{vmatrix} \]
\[ = \hat{i} (1 \cdot -1 - 0 \cdot 1) - \hat{j} (2 \cdot -1 - 1 \cdot (t + 1)) + \hat{k} (2 \cdot 0 - 1 \cdot (t + 1)) \]
\[ = -\hat{i} - (-2 - t - 1)\hat{j} - (-(t + 1))\hat{k} \]
\[ = -\hat{i} + (t + 3)\hat{j} + (t + 1)\hat{k} \]

Thus,
\[ \overrightarrow{AB} \times \overrightarrow{AC} = (-1, t + 3, t + 1) \]

### Step 3: Find the magnitude of the cross product

The area of the triangle is given by:
\[ \text{Area} = \frac{1}{2} |\overrightarrow{AB} \times \overrightarrow{AC}| \]
Calculating the magnitude:
\[ |\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{(-1)^2 + (t + 3)^2 + (t + 1)^2} \]
\[ = \sqrt{1 + (t^2 + 6t + 9) + (t^2 + 2t + 1)} \]
\[ = \sqrt{2t^2 + 8t + 11} \]

### Step 4: Find the area expression

Thus, the area of the triangle becomes:
\[ \text{Area} = \frac{1}{2} \sqrt{2t^2 + 8t + 11} \]

### Step 5: Minimize the area

To minimize the area, we need to minimize the expression under the square root:
\[ f(t) = 2t^2 + 8t + 11 \]
Taking the derivative and setting it to zero:
\[ f'(t) = 4t + 8 = 0 \implies t = -2 \]

### Step 6: Verify that this is a minimum

Taking the second derivative:
\[ f''(t) = 4 \]
Since \( f''(t) > 0 \), the function is concave up, confirming a minimum at \( t = -2 \).

### Step 7: Find the absolute value of \( t \)

Thus, the absolute value of \( t \) for which the area is minimized is:
\[ |t| = |-2| = 2 \]

### Final Answer:
The absolute value of the parameter \( t \) for which the area of the triangle is minimum is \( \boxed{2} \).