Suppose `a,b,cgt0`. If
`|(x_(1),y_(1),2a),(x_(2),y_(2),2b),(x_(3),y_(3),2c)|=abc`,
Then area of triangle whose vertices are
`A(x_(1)/a,(y_(1))/a),B((x_(2))/b,(y_(2))/b)` and `C((x_(3))/c,(y_(3))/c)` is ___

Answer

To find the area of the triangle with vertices \( A\left(\frac{x_1}{a}, \frac{y_1}{a}\right) \), \( B\left(\frac{x_2}{b}, \frac{y_2}{b}\right) \), and \( C\left(\frac{x_3}{c}, \frac{y_3}{c}\right) \), we can use the formula for the area of a triangle given by vertices in Cartesian coordinates.

### Step-by-Step Solution:

1. **Area Formula**: The area \( A \) of a triangle formed by the points \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by:
\[ A = \frac{1}{2} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right| \]

2. **Substituting the Vertices**: For our triangle, we substitute the vertices:
\[ A = \frac{1}{2} \left| \begin{vmatrix} \frac{x_1}{a} & \frac{y_1}{a} & 1 \\ \frac{x_2}{b} & \frac{y_2}{b} & 1 \\ \frac{x_3}{c} & \frac{y_3}{c} & 1 \end{vmatrix} \right| \]

3. **Simplifying the Determinant**: We can factor out \( \frac{1}{a} \), \( \frac{1}{b} \), and \( \frac{1}{c} \) from the rows:
\[ A = \frac{1}{2} \cdot \frac{1}{abc} \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right| \]

4. **Using the Given Determinant**: According to the problem statement, we know that:
\[ \left| \begin{vmatrix} x_1 & y_1 & 2a \\ x_2 & y_2 & 2b \\ x_3 & y_3 & 2c \end{vmatrix} \right| = abc \]
We can express this determinant in terms of the area:
\[ \left| \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right| = \frac{1}{2} \cdot abc \]

5. **Final Area Calculation**: Substituting this back into our area formula gives:
\[ A = \frac{1}{2} \cdot \frac{1}{abc} \cdot \frac{1}{2} \cdot abc = \frac{1}{4} \]

### Conclusion:
The area of the triangle is \( \frac{1}{4} \).