Statement -1 : If a,b,c are parameters such that `3a + 2b+4c=0,` then the family of lines `ax +by+ c =0` pass through a fixed point `(3,2).`
and
Statement -2: The equation `ax + by + c =0` wil represent a family of straight lines passing through a fixed point if there exist a linear relation between a, b and c.

Answer

To solve the problem, we will analyze both statements one by one.

### Statement 1:
"If a, b, c are parameters such that \(3a + 2b + 4c = 0\), then the family of lines \(ax + by + c = 0\) pass through a fixed point \((3,2)\)."

1. **Substituting the fixed point into the line equation:**
We need to check if the line \(ax + by + c = 0\) passes through the point \((3, 2)\). This means we substitute \(x = 3\) and \(y = 2\) into the line equation:
\[ a(3) + b(2) + c = 0 \]
This simplifies to:
\[ 3a + 2b + c = 0 \tag{1} \]

2. **Using the given condition:**
We are given that \(3a + 2b + 4c = 0\). We can rearrange this to express \(c\) in terms of \(a\) and \(b\):
\[ 4c = -3a - 2b \quad \Rightarrow \quad c = -\frac{3a + 2b}{4} \tag{2} \]

3. **Substituting equation (2) into equation (1):**
Now, we substitute \(c\) from equation (2) into equation (1):
\[ 3a + 2b - \frac{3a + 2b}{4} = 0 \]
To simplify, we can multiply the entire equation by 4 to eliminate the fraction:
\[ 4(3a + 2b) - (3a + 2b) = 0 \]
This simplifies to:
\[ 12a + 8b - 3a - 2b = 0 \]
Combining like terms gives:
\[ 9a + 6b = 0 \quad \Rightarrow \quad 3a + 2b = 0 \tag{3} \]

4. **Conclusion for Statement 1:**
From equation (3), we see that \(c\) cannot be arbitrary; it must satisfy the relation \(3a + 2b = 0\). Therefore, the line does not necessarily pass through the fixed point \((3, 2)\) for all values of \(a\), \(b\), and \(c\) satisfying \(3a + 2b + 4c = 0\). Thus, **Statement 1 is false**.

### Statement 2:
"The equation \(ax + by + c = 0\) will represent a family of straight lines passing through a fixed point if there exists a linear relation between \(a\), \(b\), and \(c\)."

1. **Understanding the linear relation:**
A linear relation among \(a\), \(b\), and \(c\) can be expressed as:
\[ ka + mb + nc = 0 \quad \text{for some constants } k, m, n \]
This implies that we can express one of the parameters in terms of the others.

2. **Finding the fixed point:**
If we set \(c = -ka - mb\), the line equation becomes:
\[ ax + by - (ka + mb) = 0 \]
Rearranging gives:
\[ ax + by + ka + mb = 0 \]
This can be rewritten as:
\[ a(x + k) + b(y + m) = 0 \]
This shows that the line passes through the point \((-k, -m)\).

3. **Conclusion for Statement 2:**
Since we can express \(c\) in terms of \(a\) and \(b\) and find a fixed point \((-k, -m)\), **Statement 2 is true**.

### Final Conclusion:
- **Statement 1 is false.**
- **Statement 2 is true.**