If `veca=a_(1)hati+a_(2)hatj+a_(3)hatk, vecb= b_(1)hati+b_(2)hatj + b_(3)hatk, vecc=c_(1)hati+c_(2)hatj+c_(3)hatk` and `[3veca+vecb \ \ 3vecb+vecc \ \ 3vecc + veca] =lambda|{:(a_1,a_2,a_3),(b_1,b_2,b_3),(c_1,c_2,c_3):}| " then find the value of " lambda/4`

Explanation

To solve the problem, we need to evaluate the scalar triple product given in the question and relate it to the determinant of the vectors provided. Let's break it down step by step.

### Step 1: Understand the Given Vectors
We have three vectors:
- \(\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}\)
- \(\vec{b} = b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}\)
- \(\vec{c} = c_1 \hat{i} + c_2 \hat{j} + c_3 \hat{k}\)

### Step 2: Write the Left-Hand Side (LHS)
We need to compute the scalar triple product:
\[ [3\vec{a} + \vec{b}, 3\vec{b} + \vec{c}, 3\vec{c} + \vec{a}] \]

### Step 3: Expand the Scalar Triple Product
Using the properties of scalar triple products, we can express this as:
\[ = (3\vec{a} + \vec{b}) \cdot ((3\vec{b} + \vec{c}) \times (3\vec{c} + \vec{a})) \]

### Step 4: Distribute the Cross Product
We can expand the cross product:
\[ (3\vec{b} + \vec{c}) \times (3\vec{c} + \vec{a}) = 3\vec{b} \times 3\vec{c} + 3\vec{b} \times \vec{a} + \vec{c} \times 3\vec{c} + \vec{c} \times \vec{a} \]
Since \(\vec{c} \times \vec{c} = \vec{0}\), we can simplify this to:
\[ = 9(\vec{b} \times \vec{c}) + 3(\vec{b} \times \vec{a}) + (\vec{c} \times \vec{a}) \]

### Step 5: Substitute Back into the Scalar Triple Product
Now substitute this back into the scalar triple product:
\[ [3\vec{a} + \vec{b}, 9(\vec{b} \times \vec{c}) + 3(\vec{b} \times \vec{a}) + (\vec{c} \times \vec{a})] \]

### Step 6: Factor Out Common Terms
We can factor out the common terms:
\[ = 3^3 [\vec{a}, \vec{b}, \vec{c}] + 3^2 [\vec{a}, \vec{b}, \vec{c}] + [\vec{a}, \vec{b}, \vec{c}] \]
This gives us:
\[ = 27[\vec{a}, \vec{b}, \vec{c}] + 9[\vec{a}, \vec{b}, \vec{c}] + [\vec{a}, \vec{b}, \vec{c}] = 37[\vec{a}, \vec{b}, \vec{c}] \]

### Step 7: Relate to the Right-Hand Side (RHS)
From the problem, we know:
\[ [3\vec{a} + \vec{b}, 3\vec{b} + \vec{c}, 3\vec{c} + \vec{a}] = \lambda |(a_1, a_2, a_3), (b_1, b_2, b_3), (c_1, c_2, c_3)| \]

### Step 8: Equate and Solve for \(\lambda\)
Setting the LHS equal to the RHS gives:
\[ 37[\vec{a}, \vec{b}, \vec{c}] = \lambda |(a_1, a_2, a_3), (b_1, b_2, b_3), (c_1, c_2, c_3)| \]
Since \([\vec{a}, \vec{b}, \vec{c}] = |(a_1, a_2, a_3), (b_1, b_2, b_3), (c_1, c_2, c_3)|\), we have:
\[ 37 = \lambda \]

### Step 9: Find \(\frac{\lambda}{4}\)
Finally, we need to find:
\[ \frac{\lambda}{4} = \frac{37}{4} = 9.25 \]

### Final Answer
Thus, the value of \(\frac{\lambda}{4}\) is:
\[ \frac{\lambda}{4} = 9.25 \]