Let `Delta_1=|{:(a_1,a_2,a_3),(b_1,b_2,b_3),(c_1,c_2,c_3):}| , Delta_2=|{:(6a_1,2a_2,2a_3),(3b_1,b_2,b_3),(12c_1, 4c_2,4c_3):}|` and `Delta_3=|{:(3a_1+b_1 , 3a_2+b_2 , 3a_3+b_3),(3b_1,3b_2,3b_3),(3c_1,3c_2,3c_3):}|`
then `Delta_3-Delta_2=kDelta_1`, find k.

Answer

To solve the problem, we need to evaluate the determinants \( \Delta_1 \), \( \Delta_2 \), and \( \Delta_3 \) and find the value of \( k \) such that \( \Delta_3 - \Delta_2 = k \Delta_1 \).

### Step 1: Evaluate \( \Delta_1 \)
Given:
\[ \Delta_1 = |(a_1, a_2, a_3), (b_1, b_2, b_3), (c_1, c_2, c_3)| \]
This determinant is simply \( \Delta_1 \).

### Step 2: Evaluate \( \Delta_2 \)
Given:
\[ \Delta_2 = |(6a_1, 2a_2, 2a_3), (3b_1, b_2, b_3), (12c_1, 4c_2, 4c_3)| \]
We can factor out constants from the rows:
- From the first row, factor out \( 2 \) (leaving \( 3a_1, a_2, a_3 \)).
- From the second row, factor out \( 3 \) (leaving \( b_1, b_2, b_3 \)).
- From the third row, factor out \( 4 \) (leaving \( 3c_1, c_2, c_3 \)).

Thus,
\[ \Delta_2 = 2 \cdot 3 \cdot 4 \cdot |(3a_1, a_2, a_3), (b_1, b_2, b_3), (3c_1, c_2, c_3)| = 24 \Delta_1 \]

### Step 3: Evaluate \( \Delta_3 \)
Given:
\[ \Delta_3 = |(3a_1 + b_1, 3a_2 + b_2, 3a_3 + b_3), (3b_1, 3b_2, 3b_3), (3c_1, 3c_2, 3c_3)| \]
We can express the first row as:
\[ (3a_1 + b_1, 3a_2 + b_2, 3a_3 + b_3) = (3a_1, 3a_2, 3a_3) + (b_1, b_2, b_3) \]
This allows us to use the property of determinants:
\[ \Delta_3 = |(3a_1, 3a_2, 3a_3), (3b_1, 3b_2, 3b_3), (3c_1, 3c_2, 3c_3)| + |(b_1, b_2, b_3), (3b_1, 3b_2, 3b_3), (3c_1, 3c_2, 3c_3)| \]
The first determinant can be factored as:
\[ 3 \cdot |(a_1, a_2, a_3), (b_1, b_2, b_3), (c_1, c_2, c_3)| = 3 \Delta_1 \]
The second determinant is zero because two rows are linearly dependent (the second row is a multiple of the first).

Thus,
\[ \Delta_3 = 3 \Delta_1 \]

### Step 4: Find \( \Delta_3 - \Delta_2 \)
Now we can find:
\[ \Delta_3 - \Delta_2 = 3 \Delta_1 - 24 \Delta_1 = (3 - 24) \Delta_1 = -21 \Delta_1 \]

### Step 5: Find \( k \)
From the equation \( \Delta_3 - \Delta_2 = k \Delta_1 \), we have:
\[ -21 \Delta_1 = k \Delta_1 \]
Thus, \( k = -21 \).

### Final Answer
The value of \( k \) is:
\[ \boxed{-21} \]