If `a+b+c=0` then the family of lines `3ax+by+2c=0` passes through the fixed point
(i) `(-2,(2)/(3))`
(ii) `(2,(3)/(3))`
(iii) `(2/3,2)`
(iv) none of these

Options

A. `(-2,(2)/(3))`
B. `(2,(3)/(3))`
C. `(2/3,2)`
D. none of these

Answer

To determine if the family of lines given by the equation \(3ax + by + 2c = 0\) passes through the fixed points when \(a + b + c = 0\), we will follow these steps:

### Step 1: Substitute \(c\) in terms of \(a\) and \(b\)

Given the equation \(a + b + c = 0\), we can express \(c\) as:
\[ c = -a - b \]

### Step 2: Substitute \(c\) into the line equation

Now, substituting \(c\) into the line equation \(3ax + by + 2c = 0\):
\[ 3ax + by + 2(-a - b) = 0 \]
This simplifies to:
\[ 3ax + by - 2a - 2b = 0 \]

### Step 3: Rearranging the equation

Rearranging the equation gives:
\[ 3ax + by = 2a + 2b \]

### Step 4: Expressing the line in slope-intercept form

We can express this in the form of \(y = mx + c\) (slope-intercept form):
\[ by = -3ax + 2a + 2b \]
Dividing through by \(b\) (assuming \(b \neq 0\)):
\[ y = -\frac{3a}{b}x + \frac{2a}{b} + 2 \]

### Step 5: Finding the fixed point

To find the fixed point, we can analyze the intercepts. The equation can be rewritten as:
\[ 3x - 2 + \frac{b}{a}(y - 2) = 0 \]
This indicates that the line passes through the point \((\frac{2}{3}, 2)\).

### Step 6: Checking the options

Now, we will check the given options to see if any of them match the point \((\frac{2}{3}, 2)\):
- (i) \((-2, \frac{2}{3})\)
- (ii) \((2, 1)\)
- (iii) \((\frac{2}{3}, 2)\)
- (iv) none of these

From our analysis, we see that option (iii) \((\frac{2}{3}, 2)\) is indeed the point through which the family of lines passes.

### Conclusion

Thus, the correct answer is:
\[ \text{(iii) } \left(\frac{2}{3}, 2\right) \]