If `z_1, z_2, z_3` be the affixes of the vertices `A, B Mand C` of a triangle having centroid at G such ;that `z = 0` is the mid point of AG then `4z_1 + Z_2 + Z_3 =`

Options

A. `4z_(1)+z_(2)+z_(3)=0`
B. `z_(1)+4z_(1)+z_(3)=0`
C. `z_(1)+z_(2)+4z_(3)=0`
D. `z_(1)+z_(2)+z_(3)=0`

Explanation

To solve the problem, we need to find the expression \(4z_1 + z_2 + z_3\) given the conditions about the triangle and its centroid.

### Step-by-Step Solution:

1. **Understanding the Centroid**:
The centroid \(G\) of a triangle with vertices \(A\), \(B\), and \(C\) has the affix given by:
\[ z_G = \frac{z_1 + z_2 + z_3}{3} \]

2. **Given Condition**:
It is given that the point \(z = 0\) is the midpoint of segment \(AG\). This means that:
\[ z = 0 = \frac{z_1 + z_G}{2} \]

3. **Substituting for \(z_G\)**:
From the expression for the centroid, we can substitute \(z_G\):
\[ 0 = \frac{z_1 + \frac{z_1 + z_2 + z_3}{3}}{2} \]

4. **Simplifying the Equation**:
Multiply both sides by 2:
\[ 0 = z_1 + \frac{z_1 + z_2 + z_3}{3} \]
Now, multiply through by 3 to eliminate the fraction:
\[ 0 = 3z_1 + z_1 + z_2 + z_3 \]
This simplifies to:
\[ 0 = 4z_1 + z_2 + z_3 \]

5. **Final Result**:
Rearranging gives us:
\[ 4z_1 + z_2 + z_3 = 0 \]

### Conclusion:
Thus, the final expression is:
\[ 4z_1 + z_2 + z_3 = 0 \]