The points `(x_(r),y_(r)),r=1,2,3` are the vertices fo an quilateral triangle of side a then the square of the determinant`D=|(x_(1),y_(1),1),(x_(2),y_(2),1),(x_(3),y_(3),1)|` equals

Options

A. `4a^(4)`
B. `3a^(4)`
C. `3/4a^(4)`
D. None of these

Answer

To solve the problem, we need to find the square of the determinant \( D = \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \) for the vertices of an equilateral triangle with side length \( a \).

### Step-by-step Solution:

1. **Understanding the Area of a Triangle**:
The area \( A \) of a triangle formed by the points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) can be calculated using the determinant:
\[ A = \frac{1}{2} \left| D \right| = \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \]

2. **Area of an Equilateral Triangle**:
The area \( A \) of an equilateral triangle with side length \( a \) is given by:
\[ A = \frac{\sqrt{3}}{4} a^2 \]

3. **Setting the Areas Equal**:
Since the area calculated using the determinant must equal the area of the equilateral triangle, we can set the two expressions for area equal to each other:
\[ \frac{1}{2} \left| D \right| = \frac{\sqrt{3}}{4} a^2 \]

4. **Solving for the Determinant**:
Rearranging the equation gives:
\[ \left| D \right| = \frac{\sqrt{3}}{2} a^2 \]

5. **Finding the Square of the Determinant**:
To find the square of the determinant \( D \), we square both sides:
\[ D^2 = \left( \frac{\sqrt{3}}{2} a^2 \right)^2 \]
\[ D^2 = \frac{3}{4} a^4 \]

### Final Result:
The square of the determinant \( D \) equals:
\[ D^2 = \frac{3}{4} a^4 \]