If `a^(2) + b^(2) + c^(3) + ab + bc + ca le 0` for all, `a, b, c in R`, then the value of the determinant
`|((a + b +2)^(2),a^(2) + b^(2),1),(1,(b +c + 2)^(2),b^(2) + c^(2)),(c^(2) + a^(2),1,(c +a +2)^(2))|`, is equal to

Options

A. `65`
B. `a^(2)+b^(2)+c^(2)+31`
C. `4(a^(2)+b^(2)+c^(2))`
D. `0`

Explanation

To solve the problem, we need to evaluate the determinant given the condition \( a^2 + b^2 + c^2 + ab + bc + ca \leq 0 \) for all \( a, b, c \in \mathbb{R} \).

### Step-by-Step Solution:

1. **Understanding the Condition**:
The expression \( a^2 + b^2 + c^2 + ab + bc + ca \) can be rewritten as:
\[ \frac{1}{2} \left( (a+b)^2 + (b+c)^2 + (c+a)^2 \right) \leq 0 \]
This implies that each squared term must be zero, leading to:
\[ a + b = 0, \quad b + c = 0, \quad c + a = 0 \]
From these equations, we can conclude that \( a = 0, b = 0, c = 0 \).

2. **Substituting Values into the Determinant**:
We substitute \( a = 0, b = 0, c = 0 \) into the determinant:
\[ D = \begin{vmatrix} (0 + 0 + 2)^2 & 0^2 + 0^2 & 1 \\ 1 & (0 + 0 + 2)^2 & 0^2 + 0^2 \\ 0^2 + 0^2 & 1 & (0 + 0 + 2)^2 \end{vmatrix} \]
This simplifies to:
\[ D = \begin{vmatrix} 4 & 0 & 1 \\ 1 & 4 & 0 \\ 0 & 1 & 4 \end{vmatrix} \]

3. **Calculating the Determinant**:
We can calculate the determinant using the formula for a 3x3 matrix:
\[ D = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our matrix:
\[ D = 4(4 \cdot 4 - 0 \cdot 1) - 0(1 \cdot 4 - 0 \cdot 0) + 1(1 \cdot 0 - 4 \cdot 0) \]
This simplifies to:
\[ D = 4(16) - 0 + 0 = 64 \]

4. **Final Calculation**:
The determinant value is:
\[ D = 64 \]

### Final Result:
Thus, the value of the determinant is \( \boxed{64} \).