If `A(z_(1))`, `B(z_(2))`, `C(z_(3))` are vertices of a triangle such that `z_(3)=(z_(2)-iz_(1))/(1-i)` and `|z_(1)|=3`, `|z_(2)|=4` and `|z_(2)+iz_(1)|=|z_(1)|+|z_(2)|`, then area of triangle `ABC` is

Options

A. `(5)/(2)`
B. `0`
C. `(25)/(2)`
D. `(25)/(4)`

Explanation

To solve the problem, we need to find the area of triangle ABC with vertices A(z1), B(z2), and C(z3) given the conditions stated. Let's break it down step by step.

### Step 1: Understanding the conditions
We are given:
- \( z_3 = \frac{z_2 - iz_1}{1 - i} \)
- \( |z_1| = 3 \)
- \( |z_2| = 4 \)
- \( |z_2 + iz_1| = |z_1| + |z_2| \)

### Step 2: Analyzing the equality of moduli
From the condition \( |z_2 + iz_1| = |z_1| + |z_2| \), we can use the triangle inequality. This equality implies that \( z_1 \) and \( z_2 \) are collinear, meaning they lie on the same line through the origin.

### Step 3: Finding the argument
Since \( z_1 \) and \( z_2 \) are collinear, we can express:
- \( z_2 = k z_1 \) for some real number \( k \).

Given \( |z_1| = 3 \) and \( |z_2| = 4 \), we have:
\[ |k| \cdot |z_1| = 4 \implies |k| \cdot 3 = 4 \implies |k| = \frac{4}{3} \]
Thus, \( z_2 = \frac{4}{3} z_1 \) or \( z_2 = -\frac{4}{3} z_1 \).

### Step 4: Finding \( z_3 \)
Using \( z_2 = \frac{4}{3} z_1 \), we can substitute into the equation for \( z_3 \):
\[ z_3 = \frac{\frac{4}{3} z_1 - iz_1}{1 - i} \]
Simplifying the numerator:
\[ z_3 = \frac{z_1 \left( \frac{4}{3} - i \right)}{1 - i} \]
Now, we can multiply the numerator and denominator by the conjugate of the denominator:
\[ z_3 = \frac{z_1 \left( \frac{4}{3} - i \right)(1 + i)}{(1 - i)(1 + i)} = \frac{z_1 \left( \frac{4}{3} + \frac{4}{3}i - i - 1 \right)}{2} = \frac{z_1 \left( \frac{4}{3} - 1 + \left( \frac{4}{3} - 1 \right)i \right)}{2} \]
Calculating \( \frac{4}{3} - 1 = \frac{1}{3} \):
\[ z_3 = \frac{z_1 \left( \frac{1}{3} + \left( \frac{1}{3} \right)i \right)}{2} = \frac{z_1}{6} (1 + i) \]

### Step 5: Area of triangle ABC
The area of triangle ABC can be computed using the formula:
\[ \text{Area} = \frac{1}{2} \times |z_1| \times |z_2| \times \sin(\theta) \]
where \( \theta \) is the angle between \( z_1 \) and \( z_2 \). Since \( z_1 \) and \( z_2 \) are perpendicular (as shown earlier), \( \sin(\theta) = 1 \).

Thus, substituting the values:
\[ \text{Area} = \frac{1}{2} \times 3 \times 4 \times 1 = 6 \]

### Step 6: Final area calculation
Now we can calculate the area using the coordinates of points A, B, and C:
- The lengths of sides can be derived from the coordinates, but since we already established the perpendicularity, we can simplify:
\[ \text{Area} = \frac{1}{2} \times BC \times AC \]
Since both lengths are derived from the magnitudes of \( z_1 \) and \( z_2 \):
\[ \text{Area} = \frac{1}{2} \times 3 \times 4 = 6 \]

### Conclusion
The area of triangle ABC is \( 6 \).