If the vectors `veca=hati+ahatj+a^(2)hatk, vecb=hati+bhatj+b^(2)hatk, vecc=hati+chatj+c^(2)hatk` are three non-coplanar vectors and `| (a, a^(2), 1+a^(3)),(b,b^(2),1+b^(3)),(c,c^(2),1+c^(3))|=0` , then the value of `abc` is

Options

A. 0
B. 1
C. 2
D. `-1`

Explanation

To solve the given problem, we need to analyze the condition provided for the vectors \( \vec{a}, \vec{b}, \vec{c} \) and the determinant condition. Let's break down the solution step by step.

### Step 1: Write down the vectors
The vectors are given as:
\[ \vec{a} = \hat{i} + a\hat{j} + a^2\hat{k} \]
\[ \vec{b} = \hat{i} + b\hat{j} + b^2\hat{k} \]
\[ \vec{c} = \hat{i} + c\hat{j} + c^2\hat{k} \]

### Step 2: Formulate the determinant
We need to analyze the determinant:
\[ \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \]
This determinant represents the condition for the vectors to be non-coplanar. For non-coplanarity, this determinant must not equal zero.

### Step 3: Calculate the determinant
Calculating the determinant, we have:
\[ = 1 \cdot (b \cdot b^2 - c \cdot b^2) - a \cdot (1 \cdot b^2 - 1 \cdot c^2) + a^2 \cdot (1 \cdot c - 1 \cdot b) \]
This simplifies to:
\[ = (b - c)(b^2 - a^2) + (c - b)(a^2 - b^2) + (a - b)(c - a) \]
This determinant must not equal zero for the vectors to be non-coplanar.

### Step 4: Analyze the given condition
We are also given that:
\[ \begin{vmatrix} 1 & a^2 & 1 + a^3 \\ 1 & b^2 & 1 + b^3 \\ 1 & c^2 & 1 + c^3 \end{vmatrix} = 0 \]
This determinant equals zero, which implies that the vectors are coplanar in this context.

### Step 5: Simplify the determinant
Using row operations, we can simplify the determinant:
1. Subtract the first row from the second and third rows:
\[ \begin{vmatrix} 1 & a^2 & 1 + a^3 \\ 0 & b^2 - a^2 & b^3 - a^3 \\ 0 & c^2 - a^2 & c^3 - a^3 \end{vmatrix} \]
2. This leads to:
\[ = (b^2 - a^2)(c^3 - a^3) - (c^2 - a^2)(b^3 - a^3) \]
3. Factor out the differences:
\[ = (b - a)(b + a)(c - a)(c + a) = 0 \]

### Step 6: Solve for \( abc \)
Since the determinant is zero, we can conclude that:
\[ 1 + abc = 0 \]
Thus, we find:
\[ abc = -1 \]

### Final Answer
The value of \( abc \) is:
\[ \boxed{-1} \]