Find the value of `lambda` for which `|{:(2a_1+b_1 , 2a_2+b_2 , 2a_3+b_3),(2b_1+c_1, 2b_2+c_2 , 2b_3+c_3),(2c_1+a_1,2c_2+a_2, 2c_3+a_3):}|=lambda|{:(a_1,a_2,a_3),(b_1,b_2,b_3),(c_1,c_2,c_3):}|`

Answer

To find the value of \(\lambda\) for which

\[ \left| \begin{array}{ccc} 2a_1 + b_1 & 2a_2 + b_2 & 2a_3 + b_3 \\ 2b_1 + c_1 & 2b_2 + c_2 & 2b_3 + c_3 \\ 2c_1 + a_1 & 2c_2 + a_2 & 2c_3 + a_3 \end{array} \right| = \lambda \left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right| \]

we will follow these steps:

### Step 1: Write the Determinant
The given determinant can be expressed as:

\[ D = \left| \begin{array}{ccc} 2a_1 + b_1 & 2a_2 + b_2 & 2a_3 + b_3 \\ 2b_1 + c_1 & 2b_2 + c_2 & 2b_3 + c_3 \\ 2c_1 + a_1 & 2c_2 + a_2 & 2c_3 + a_3 \end{array} \right| \]

### Step 2: Factor Out Common Terms
We can factor out a 2 from each column of the determinant:

\[ D = 2 \left| \begin{array}{ccc} a_1 + \frac{1}{2}b_1 & a_2 + \frac{1}{2}b_2 & a_3 + \frac{1}{2}b_3 \\ b_1 + \frac{1}{2}c_1 & b_2 + \frac{1}{2}c_2 & b_3 + \frac{1}{2}c_3 \\ c_1 + \frac{1}{2}a_1 & c_2 + \frac{1}{2}a_2 & c_3 + \frac{1}{2}a_3 \end{array} \right| \]

### Step 3: Rearranging the Determinant
Now we can rearrange the terms in the determinant. We can express it as a sum of two determinants:

\[ D = 2 \left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right| + \text{(other terms)} \]

### Step 4: Use Properties of Determinants
The determinant can also be simplified using properties of determinants. We can express \(D\) as:

\[ D = 2^3 \left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right| \]

### Step 5: Comparing Both Sides
Now, we can compare both sides of the equation:

\[ D = 8 \left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right| = \lambda \left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right| \]

### Step 6: Solve for \(\lambda\)
From the comparison, we get:

\[ 8 = \lambda \]

### Final Result
Thus, the value of \(\lambda\) is:

\[ \lambda = 8 \]