If `vec(a_(1)),vec(a_(2)),vec(a_(3)) and vec(b_(1)), vec(b_(2)),vec(b_(3))` be two sets of non - coplanar vectors, such that `vec(a_(p)).vec(b_(q))={{:(0,"if",pneq),(4,"if",p=q):}" for "p=1,2,3 and q=1,2,3`, then the value of `[vec(a_(1))2vec(a_(2))3vec(a_(3))][(vec(b_(1))+vec(b_(2)),vec(b_(2))+vec(b_(3)),vec(b_(3)),vec(b_(3))+vec(b_(1)))]` is equal to

Answer

To solve the given problem, we will follow a systematic approach to evaluate the expression involving the non-coplanar vectors \(\vec{a_1}, \vec{a_2}, \vec{a_3}\) and \(\vec{b_1}, \vec{b_2}, \vec{b_3}\).

### Step-by-Step Solution:

1. **Understanding the Dot Product Conditions**:
We have two sets of vectors \(\vec{a_1}, \vec{a_2}, \vec{a_3}\) and \(\vec{b_1}, \vec{b_2}, \vec{b_3}\). The problem states that:
\[ \vec{a_p} \cdot \vec{b_q} = \begin{cases} 0 & \text{if } p \neq q \\ 4 & \text{if } p = q \end{cases} \]
This means that the dot product is 4 when the indices are the same and 0 otherwise.

2. **Setting Up the Expression**:
We need to evaluate the expression:
\[ [\vec{a_1} \, 2\vec{a_2} \, 3\vec{a_3}] \begin{pmatrix} \vec{b_1} + \vec{b_2} \\ \vec{b_2} + \vec{b_3} \\ \vec{b_3} + \vec{b_1} \end{pmatrix} \]
This can be rewritten as:
\[ 2\vec{a_2} \cdot (\vec{b_1} + \vec{b_2}) + 3\vec{a_3} \cdot (\vec{b_2} + \vec{b_3}) + \vec{a_1} \cdot (\vec{b_3} + \vec{b_1}) \]

3. **Calculating Each Dot Product**:
- For \(\vec{a_1} \cdot (\vec{b_3} + \vec{b_1})\):
\[ \vec{a_1} \cdot \vec{b_3} + \vec{a_1} \cdot \vec{b_1} = 0 + 4 = 4 \]
- For \(2\vec{a_2} \cdot (\vec{b_1} + \vec{b_2})\):
\[ 2(\vec{a_2} \cdot \vec{b_1} + \vec{a_2} \cdot \vec{b_2}) = 2(0 + 4) = 8 \]
- For \(3\vec{a_3} \cdot (\vec{b_2} + \vec{b_3})\):
\[ 3(\vec{a_3} \cdot \vec{b_2} + \vec{a_3} \cdot \vec{b_3}) = 3(4 + 0) = 12 \]

4. **Summing Up the Results**:
Now, we add all the calculated values:
\[ 4 + 8 + 12 = 24 \]

5. **Final Calculation**:
The expression we need to evaluate is:
\[ 12 \times 24 = 288 \]

### Conclusion:
The value of the expression is \(288\).